【剑指Offer】Java题解汇总
【剑指Offer】Java题解汇总
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- 01. 数组中重复的数字
- 02. 二维数组中的查找
- 03. 替换空格
- 04. 从尾到头打印链表
- 05. 重建二叉树
- 06. 用两个栈实现队列
- 07. 斐波那契数列
- 08. 青蛙跳台阶问题
- 09. 旋转数组的最小数字
- 10. 矩阵中的路径
- 11. 机器人的运动范围
- 12. 剪绳子<动态规划>
- 13. 剪绳子<贪心算法，大数取余>
- 14. 二进制中1的个数
- 15. 数值的整数次方
- 16. 打印从1到最大的n位数（大数问题）
- 17. 删除链表的节点
- 18. 正则表达式匹配（Hard）
- 19. 表示数值的字符串
- 20. 调整数组顺序使奇数位于偶数前面
- 21. 链表中倒数第k个节点
- 22. 反转链表
- 23. 合并两个排序的链表
- 24. 树的子结构
- 25. 二叉树的镜像
- 26. 对称的二叉树
- 27. 顺时针打印矩阵
- 28. 包含min函数的栈
- 29. 栈的压入、弹出序列
- 30. 从上到下打印二叉树（1）
- 31. 从上到下打印二叉树（2）
- 32. 从上到下打印二叉树（3）之字形
- 33. 二叉搜索树的后序遍历序列
- 34. 二叉树中和为某一值的路径
- 35. 复杂链表的复制
- 36. 二叉搜索树与双向链表
- 37. 序列化/反序列化二叉树
- 38. 字符串的排列
- 39. 数组中出现次数超过一半的数字
- 40. 最小的k个数
- 41. 数据流中的中位数
- 42. 连续子数组的最大和
- 43. 1-n整数中1出现的次数
- 44. 数字序列中某一位的数字
- 45. 把数组排成最小的数
- 46. 把数字翻译成字符串
- 47. 礼物的最大价值
- 48. 最长不含重复字符的子字符串
- 49. 丑数
- 50. 第一个只出现一次的字符
- 51. 数组中的逆序对
- 52. 两个链表的第一个公共节点
- 53. 在排序数组中查找数字（1）
- 54. 在排序数组中查找数字（2）
- 55. 二叉搜索树的第k大节点
- 56. 二叉树的深度
- 57. 平衡二叉树
- 58. 数组中数字出现的次数（1）
- 59. 数组中数字出现的次数（2）
- 60. 和为s的两个数字
- 61. 和为s的连续正数序列
- 62. 翻转单词顺序
- 63. 左旋转字符串
- 64. 滑动窗口的最大值
- 65. 队列的最大值
- 66. 扑克牌中的顺子
- 67. 圆圈中最后剩下的数字
- 68. 股票的最大利润
- 69. 求1+2+。。。+n
- 70. 不用加减乘除做加法
- 71. 构建乘积数组
- 72. 把字符串转换为整数
- 73. 二叉搜索树的最近公共祖先
- 74. 二叉树的最近公共祖先
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01. 数组中重复的数字
class Solution {
public int findRepeatNumber(int[] nums) {
for(int i = 0;i < nums.length; i++){
while(nums[i] != i){
if(nums[i] == nums[nums[i]]){
return nums[i];
} else {
int temp = nums[i];
nums[i] = nums[temp];
nums[temp] = temp;
}
}
}
return -1;
}
}
02. 二维数组中的查找
class Solution {
public boolean findNumberIn2DArray(int[][] matrix, int target) {
if(matrix.length == 0){
return false;
}
int row = 0;
int col = matrix[0].length-1;
while(row < matrix.length && col >= 0){
if(target == matrix[row][col]){
return true;
} else if(target > matrix[row][col]) {
row ++;
} else if(target < matrix[row][col]) {
col --;
}
}
return false;
}
}
03. 替换空格
// class Solution {
// public String replaceSpace(String s) {
// return s.replaceAll(" ","%20");
// }
// }
class Solution{
public String replaceSpace(String s) {
if(s == null){
return null;
}
int cnt = 0;
for(int i = 0;i < s.length();i++){
if(s.charAt(i) == ' '){
cnt ++;
}
}
cnt = s.length() + cnt * 2;
char[] s_char = new char[cnt];
int p1 = s.length()-1;
int p2 = cnt-1;
while(p1>=0 && p2>=0){
if(s.charAt(p1) != ' '){
s_char[p2] = s.charAt(p1);
p2 --;
p1 --;
} else{
s_char[p2] = '0';
p2 --;
s_char[p2] = '2';
p2 --;
s_char[p2] = '%';
p2 --;
p1 --;
}
}
return new String(s_char);
}
}
04. 从尾到头打印链表
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
//用栈的方法
class Solution {
public int[] reversePrint(ListNode head) {
if(head == null){
return new int[]{
};
}
ListNode temp = head;
Stack<ListNode> stack = new Stack<ListNode>();
while(temp != null){
stack.push(temp);
temp = temp.next;
}
int size = stack.size();
int[] vec = new int[size];
for(int i = 0;i < size;i++){
vec[i] = stack.pop().val;
}
return vec;
}
}
05. 重建二叉树
官方：
方法一：递归
二叉树的前序遍历顺序是：根节点、左子树、右子树，每个子树的遍历顺序同样满足前序遍历顺序。
二叉树的中序遍历顺序是：左子树、根节点、右子树，每个子树的遍历顺序同样满足中序遍历顺序。
前序遍历的第一个节点是根节点，只要找到根节点在中序遍历中的位置，在根节点之前被访问的节点都位于左子树，在根节点之后被访问的节点都位于右子树，由此可知左子树和右子树分别有多少个节点。
由于树中的节点数量与遍历方式无关，通过中序遍历得知左子树和右子树的节点数量之后，可以根据节点数量得到前序遍历中的左子树和右子树的分界，因此可以进一步得到左子树和右子树各自的前序遍历和中序遍历，可以通过递归的方式，重建左子树和右子树，然后重建整个二叉树。
使用一个 Map 存储中序遍历的每个元素及其对应的下标，目的是为了快速获得一个元素在中序遍历中的位置。调用递归方法，对于前序遍历和中序遍历，下标范围都是从 0 到 n-1，其中 n 是二叉树节点个数。
递归方法的基准情形有两个：判断前序遍历的下标范围的开始和结束，若开始大于结束，则当前的二叉树中没有节点，返回空值 null。若开始等于结束，则当前的二叉树中恰好有一个节点，根据节点值创建该节点作为根节点并返回。
若开始小于结束，则当前的二叉树中有多个节点。在中序遍历中得到根节点的位置，从而得到左子树和右子树各自的下标范围和节点数量，知道节点数量后，在前序遍历中即可得到左子树和右子树各自的下标范围，然后递归重建左子树和右子树，并将左右子树的根节点分别作为当前根节点的左右子节点。
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public TreeNode buildTree(int[] preorder, int[] inorder) {
if (preorder == null || preorder.length == 0) {
return null;
}
Map<Integer, Integer> indexMap = new HashMap<Integer, Integer>();
int length = preorder.length;
for (int i = 0; i < length; i++) {
indexMap.put(inorder[i], i);
}
TreeNode root = buildTree(preorder, 0, length - 1, inorder, 0, length - 1, indexMap);
return root;
}
public TreeNode buildTree(int[] preorder, int preorderStart, int preorderEnd, int[] inorder, int inorderStart, int inorderEnd, Map<Integer, Integer> indexMap) {
if (preorderStart > preorderEnd) {
return null;
}
int rootVal = preorder[preorderStart];
TreeNode root = new TreeNode(rootVal);
if (preorderStart == preorderEnd) {
return root;
} else {
int rootIndex = indexMap.get(rootVal);
int leftNodes = rootIndex - inorderStart, rightNodes = inorderEnd - rootIndex;
TreeNode leftSubtree = buildTree(preorder, preorderStart + 1, preorderStart + leftNodes, inorder, inorderStart, rootIndex - 1, indexMap);
TreeNode rightSubtree = buildTree(preorder, preorderEnd - rightNodes + 1, preorderEnd, inorder, rootIndex + 1, inorderEnd, indexMap);
root.left = leftSubtree;
root.right = rightSubtree;
return root;
}
}
}
我的：
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public TreeNode buildTree(int[] preorder, int[] inorder) {
if(preorder == null || preorder.length == 0){
return null;
}
int length = preorder.length;
Map<Integer,Integer> indexMap = new HashMap<Integer,Integer>();
for(int i=0; i<length; i++){
indexMap.put(inorder[i],i);
}
TreeNode root = rebuildTree(preorder,0,length-1,inorder,0,length-1,indexMap);
return root;
}
public TreeNode rebuildTree(int[] preorder,int start_pre,int end_pre,int[] inorder,int start_in,int end_in,Map<Integer,Integer> indexMap){
if(start_pre > end_pre){
return null;
}
int rootVal = preorder[start_pre];
TreeNode root = new TreeNode(rootVal);
if(start_pre == end_pre){
return root;
} else {
int index = indexMap.get(root.val);
int leftNodes = index - start_in;
int rightNodes = end_in - index;
TreeNode leftTree = rebuildTree(preorder,start_pre+1,start_pre+leftNodes,inorder,start_in,index-1,indexMap);
TreeNode rightTree = rebuildTree(preorder,start_pre+leftNodes+1,end_pre,inorder,index+1,end_in,indexMap);
root.left = leftTree;
root.right = rightTree;
return root;
}
}
}
06. 用两个栈实现队列
class CQueue {
Stack<Integer> stack1;
Stack<Integer> stack2;
public CQueue() {
stack1 = new Stack<Integer>();
stack2 = new Stack<Integer>();
}
public void appendTail(int value) {
stack1.push(value);
}
public int deleteHead() {
if(!stack2.empty()){
return stack2.pop();
} else {
if(stack1.empty()){
return -1;
} else{
while(!stack1.empty()){
stack2.push(stack1.pop());
}
return stack2.pop();
}
}
}
}
/**
* Your CQueue object will be instantiated and called as such:
* CQueue obj = new CQueue();
* obj.appendTail(value);
* int param_2 = obj.deleteHead();
*/
07. 斐波那契数列
class Solution {
public int fib(int n) {
if(n == 0){
return 0;
}
if(n == 1){
return 1;
}
int f0 = 0;
int f1 = 1;
int fn = 0;
for(int i=1; i<n;i++){
fn = (f0 + f1) % 1000000007;
f0 = f1;
f1 = fn;
}
return fn;
}
}
08. 青蛙跳台阶问题
类似于斐波那契数列
class Solution {
public int numWays(int n) {
if(n == 1 || n==0){
return 1;
}
if(n == 2){
return 2;
}
int fn_1 = 1;
int fn_2 = 2;
int fn_3 = 0;
for(int i=2; i<n; i++){
fn_3 = (fn_1+fn_2) % 1000000007;
fn_1 = fn_2;
fn_2 = fn_3;
}
return fn_3;
}
}
09. 旋转数组的最小数字
class Solution {
public int minArray(int[] numbers) {
int index1 = 0;
int index2 = numbers.length-1;
int indexMid;
if(numbers[index1] < numbers[index2]){
return numbers[index1];
}
while((index2-index1) > 1){
indexMid = (index1+index2)/2;
//如果index1和index2和indexMid的值都一样，就只能用顺序查找了
if(numbers[index1] == numbers[index2] && numbers[index2] == numbers[indexMid]){
int res = numbers[index1];
for(int i=index1+1; i<=index2; i++){
if(res > numbers[i]){
res = numbers[i];
}
}
return res;
}
if(numbers[indexMid] >= numbers[index1]){
index1 = indexMid;
} else if(numbers[indexMid] <= numbers[index2]){
index2 = indexMid;
}
}
return numbers[index2];
}
}
10. 矩阵中的路径
还有一些问题：
class Solution {
public boolean exist(char[][] board, String word) {
int rows = board.length;
int cols = board[0].length;
int wordLength = word.length();
if(wordLength>(rows*cols)){
return false;
}
boolean[][] map = new boolean[rows][cols];
boolean flag = false;;
for(int row=0; row<rows; row++){
for(int col=0; col<cols; col++){
if(board[row][col] == word.charAt(0)){
flag = hasPath(board,row,col,map,0,word);
}
if(flag == true){
return true;
}
}
}
return false;
}
public boolean hasPath(char[][] board, int i, int j, boolean[][] map, int wordIndex, String word){
if((wordIndex+1) == word.length()){
return true;
} else {
if(i>=0 && i<board.length && j>=0 && j<board[0].length && map[i][j] == false && board[i][j]==word.charAt(wordIndex)){
map[i][j] = true;
wordIndex ++;
if(hasPath(board,i+1,j,map,wordIndex,word)){
return true;
} else if(hasPath(board,i-1,j,map,wordIndex,word)){
return true;
} else if(hasPath(board,i,j+1,map,wordIndex,word)){
return true;
} else if(hasPath(board,i,j-1,map,wordIndex,word)){
return true;
} else{
return false;
}
} else{
return false;
}
}
}
}
11. 机器人的运动范围
class Solution {
public int movingCount(int m, int n, int k) {
boolean[] visited = new boolean[m*n];
int cnt = mcount(m,n,k,0,0,visited);
return cnt;
}
public int mcount(int m, int n, int k, int row, int col, boolean[] visited){
int cnt = 0;
if(check(m,n,k,row,col,visited)){
visited[row*n+col] = true;
cnt = 1+mcount(m,n,k,row-1,col,visited) + mcount(m,n,k,row+1,col,visited) + mcount(m,n,k,row,col-1,visited) + mcount(m,n,k,row,col+1,visited);
}
return cnt;
}
public boolean check(int m, int n, int k, int row, int col, boolean[] visited){
if(row>=0 && row<m && col>=0 && col<n && getSum(row,col)<=k && !visited[row*n+col]){
return true;
}
return false;
}
public int getSum(int row, int col){
int sum = 0;
while(row > 0){
sum = sum + row%10;
row = row/10;
}
while(col > 0){
sum = sum + col%10;
col = col/10;
}
return sum;
}
}
12. 剪绳子<动态规划>
class Solution {
public int cuttingRope(int n) {
if(n < 2){
return 0;
}
if(n == 2){
return 1;
}
if(n == 3){
return 2;
}
int[] products = new int[n+1];
products[0] = 0;
//这里的原因是，如果长度是1、2、3时就不要切分了，因为分了以后更小了
products[1] = 1;
products[2] = 2;
products[3] = 3;
int max = 0;
for(int i=4; i<=n; i++){
max = 0;
for(int j=1; j<=i/2; j++){
int product = products[j]*products[i-j];
if(max < product){
max = product;
}
}
products[i] = max;
}
return products[n];
}
}
13. 剪绳子<贪心算法，大数取余>
参考详解：https://leetcode-cn.com/problems/jian-sheng-zi-ii-lcof/solution/mian-shi-ti-14-ii-jian-sheng-zi-iitan-xin-er-fen-f/
class Solution {
public int cuttingRope(int n) {
if(n < 2){
return 0;
}
if(n == 2){
return 1;
}
if(n == 3){
return 2;
}
//尽可能多的减去长度为3的绳子段
int time3 = n/3;
if(n - time3*3 == 1){
time3 = time3-1;
}
int time2 = (n-time3*3)/2;
long res = 1;
//循环取余
for(int i=0; i<time3; i++){
res = (res * 3) % 1000000007;
}
for(int i=0; i<time2; i++){
res = (res * 2) % 1000000007;
}
// long res = (long)(Math.pow(3,time3)) * (long)(Math.pow(2,time2)) % 1000000007;
return (int)res;
}
}
14. 二进制中1的个数
Java提供的位运算符有：左移( << )、右移( >> ) 、无符号右移( >>> ) 、位与( & ) 、位或( | )、位非( ~ )、位异或( ^ )
方法一：
public class Solution {
// you need to treat n as an unsigned value
public int hammingWeight(int n) {
int count = 0;
while(n != 0){
if((n&1) == 1){
count ++;
}
n= n>>>1;
}
return count;
}
}
牛皮方法二：
public class Solution {
// you need to treat n as an unsigned value
public int hammingWeight(int n) {
/*
int count = 0;
while(n != 0){
if((n&1) == 1){
count ++;
}
n= n>>>1;
}
return count;
*/
int count = 0;
while(n != 0){
count ++;
n = (n-1)&n;
}
return count;
}
}
15. 数值的整数次方
参考大佬：https://leetcode-cn.com/problems/shu-zhi-de-zheng-shu-ci-fang-lcof/solution/mian-shi-ti-16-shu-zhi-de-zheng-shu-ci-fang-kuai-s/
/*class Solution {
public double myPow(double x, int n) {
if((x-0.0) < 0.00000001 && n < 0){
return 0.0;
}
if(n == 0){
return 1.0;
}
if(n < 0){
double res = 1.0 / powerWithPositiveEx(x,-n);
return res;
}
double res = powerWithPositiveEx(x,n);
return res;
}
/*普通做法
public double powerWithPositiveEx(double x, int n){
double res = 1.0;
for(int i=0; i/*
public double powerWithPositiveEx(double x, int n){
int cnt = (int)(Math.log(n)/Math.log(2)) - 1;
double res = x;
for(int i=1; i
class Solution {
public double myPow(double x, int n) {
if(x == 0) return 0;
long b = n;
double res = 1.0;
if(b < 0) {
x = 1 / x;
b = -b;
}
while(b > 0) {
if((b & 1) == 1) res *= x;
x *= x;
b >>= 1;
}
return res;
}
}
16. 打印从1到最大的n位数（大数问题）
class Solution {
int res[];
int index = 0;
public int[] printNumbers(int n) {
if(n <= 0){
return new int[]{
};
}
char[] number = new char[n];
res = new int[(int)Math.pow(10, n) - 1];
for(int i=0; i<10; i++){
number[0] = (char)(i+'0');
printToMax(number,n,0);
}
return res;
}
public void printToMax(char[] number, int length, int index){
if(index == length-1){
printIntoInt(number);
return;
}
for(int i=0; i<10; i++){
number[index+1] = (char)(i+'0');
printToMax(number,length,index+1);
}
}
public void printIntoInt(char[] number){
int nLength = number.length;
int result = 0;
int j = 1;
for(int i=nLength-1; i>=0; i--,j*=10){
result = result + (number[i]-'0')*j;
}
if(result != 0){
res[index] = result;
index ++;
}
}
}
17. 删除链表的节点
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode deleteNode(ListNode head, int val) {
//没有节点
if(head == null){
return head;
}
//只有一个节点，正好要删除他
if(head.next == null && head.val == val){
return null;
}
//要删除头节点
if(head.val == val){
return head.next;
}
ListNode temp = head;
while(temp != null && temp.next != null){
if(temp.next.val == val){
temp.next = temp.next.next;
}
temp = temp.next;
}
return head;
}
}
18. 正则表达式匹配（Hard）
有递归和动态规划两种方法，这里只有递归
/*有问题
class Solution {
public boolean isMatch(String s, String p) {
if(s == null || p == null){
return false;
}
int sIndex = 0;
int pIndex = 0;
return matchCore(s,p,sIndex,pIndex);
}
public boolean matchCore(String s, String p, int sIndex, int pIndex){
if(sIndex >= s.length()-1 && pIndex >= p.length()-1){
return true;
}
if(sIndex != s.length()-1 && pIndex == p.length()-1){
return false;
}
if((pIndex+1)
class Solution {
public boolean isMatch(String A, String B) {
// 如果字符串长度为0，需要检测下正则串
if (A.length() == 0) {
// 如果正则串长度为奇数，必定不匹配，比如 "."、"ab*",必须是 a*b*这种形式，*在奇数位上
if (B.length() % 2 != 0) return false;
int i = 1;
while (i < B.length()) {
if (B.charAt(i) != '*') return false;
i += 2;
}
return true;
}
// 如果字符串长度不为0，但是正则串没了，return false
if (B.length() == 0) return false;
// c1 和 c2 分别是两个串的当前位，c3是正则串当前位的后一位，如果存在的话，就更新一下
char c1 = A.charAt(0), c2 = B.charAt(0), c3 = 'a';
if (B.length() > 1) {
c3 = B.charAt(1);
}
// 和dp一样，后一位分为是 '*' 和不是 '*' 两种情况
if (c3 != '*') {
// 如果该位字符一样，或是正则串该位是 '.',也就是能匹配任意字符，就可以往后走
if (c1 == c2 || c2 == '.') {
return isMatch(A.substring(1), B.substring(1));
} else {
// 否则不匹配
return false;
}
} else {
// 如果该位字符一样，或是正则串该位是 '.'，和dp一样，有看和不看两种情况
if (c1 == c2 || c2 == '.') {
return isMatch(A.substring(1), B) || isMatch(A, B.substring(2));
} else {
// 不一样，那么正则串这两位就废了，直接往后走
return isMatch(A, B.substring(2));
}
}
}
}
19. 表示数值的字符串
class Solution {
private int index = 0;//全局索引
private boolean scanUnsignedInteger(String str) {
//是否包含无符号数
int before = index;
while(str.charAt(index) >= '0' && str.charAt(index) <= '9')
index++;
return index > before;
}
private boolean scanInteger(String str) {
//是否包含有符号数
if(str.charAt(index) == '+' || str.charAt(index) == '-')
index++;
return scanUnsignedInteger(str);
}
public boolean isNumber(String s) {
//空字符串
if(s == null || s.length() == 0)
return false;
//添加结束标志
s = s + '|';
//跳过首部的空格
while(s.charAt(index) == ' ')
index++;
boolean numeric = scanInteger(s); //是否包含整数部分
if(s.charAt(index) == '.') {
index++;
//有小数点，处理小数部分
//小数点两边只要有一边有数字就可以，所以用||，
//注意scanUnsignedInteger要在前面，否则不会进
numeric = scanUnsignedInteger(s) || numeric;
}
if((s.charAt(index) == 'E' || s.charAt(index) == 'e')) {
index++;
//指数部分
//e或E的两边都要有数字，所以用&&
numeric = numeric && scanInteger(s);
}
//跳过尾部空格
while(s.charAt(index) == ' ')
index++;
return numeric && s.charAt(index) == '|' ;
}
}
20. 调整数组顺序使奇数位于偶数前面
class Solution {
public int[] exchange(int[] nums) {
if(nums.length == 0 ||nums.length == 1){
return nums;
}
int indexOne = 0;
int indexTwo = nums.length-1;
while(indexTwo > indexOne){
while(indexOne < indexTwo && nums[indexOne]%2 != 0){
indexOne ++;
}
while(indexTwo > indexOne && nums[indexTwo]%2 != 1){
indexTwo --;
}
if(indexTwo <= indexOne){
break;
}
int temp = nums[indexOne];
nums[indexOne] = nums[indexTwo];
nums[indexTwo] = temp;
indexOne ++;
indexTwo --;
}
return nums;
}
}
21. 链表中倒数第k个节点
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode getKthFromEnd(ListNode head, int k) {
if(head == null || k == 0){
return null;
}
ListNode tempOne = head;
ListNode tempTwo = head;
for(int i=0; i<k-1; i++){
if(tempOne.next != null){
tempOne = tempOne.next;
} else {
return null;
}
}
while(tempOne.next != null){
tempOne = tempOne.next;
tempTwo = tempTwo.next;
}
return tempTwo;
}
}
22. 反转链表
用的是一个个嵌进去的方法，参考尚学堂数据结构课
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode reverseList(ListNode head) {
if(head == null || head.next == null){
return head;
}
ListNode reverseHead = new ListNode(-1);
ListNode tempOri = head;
ListNode tempNext;
while(tempOri != null){
tempNext = tempOri.next;
tempOri.next = reverseHead.next;
reverseHead.next = tempOri;
if(tempNext == null){
break;
}
tempOri = tempNext;
tempNext = tempNext.next;
}
return reverseHead.next;
}
}
23. 合并两个排序的链表
循环迭代：
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
if(l1 == null){
return l2;
}
if(l2 == null){
return l1;
}
ListNode preHead = new ListNode(-1);
ListNode temp = preHead;
while(l1 != null && l2 != null){
if(l1.val <= l2.val){
temp.next = l1;
temp = temp.next;
l1 = l1.next;
} else {
temp.next = l2;
temp = temp.next;
l2 = l2.next;
}
}
if(l1 == null){
temp.next = l2;
} else{
temp.next = l1;
}
return preHead.next;
}
}
递归：
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
/*
if(l1 == null){
return l2;
}
if(l2 == null){
return l1;
}
ListNode preHead = new ListNode(-1);
ListNode temp = preHead;
while(l1 != null && l2 != null){
if(l1.val <= l2.val){
temp.next = l1;
temp = temp.next;
l1 = l1.next;
} else {
temp.next = l2;
temp = temp.next;
l2 = l2.next;
}
}
if(l1 == null){
temp.next = l2;
} else{
temp.next = l1;
}
return preHead.next;
*/
if(l1 == null){
return l2;
}
if(l2 == null){
return l1;
}
if(l1.val <= l2.val){
l1.next = mergeTwoLists(l1.next,l2);
return l1;
} else {
l2.next = mergeTwoLists(l1,l2.next);
return l2;
}
}
}
24. 树的子结构
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public boolean isSubStructure(TreeNode A, TreeNode B) {
boolean res = false;
if(A != null && B != null){
if(A.val == B.val){
res = proveSubStructure(A,B);
}
if(res == false){
res = isSubStructure(A.left,B);
}
if(res == false){
res = isSubStructure(A.right,B);
}
}
return res;
}
public boolean proveSubStructure(TreeNode A, TreeNode B){
if(B == null){
return true;
}
if(A == null){
return false;
}
if(A.val != B.val){
return false;
}
return proveSubStructure(A.left,B.left) && proveSubStructure(A.right,B.right);
}
}
25. 二叉树的镜像
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public TreeNode mirrorTree(TreeNode root) {
if(root == null){
return root;
}
if(root.left == null && root.right ==null){
return root;
}
TreeNode temp = root.left;
root.left = root.right;
root.right = temp;
if(root.left != null){
mirrorTree(root.left);
}
if(root.right != null){
mirrorTree(root.right);
}
return root;
}
}
26. 对称的二叉树
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public boolean isSymmetric(TreeNode root) {
return isSym(root,root);
}
public boolean isSym(TreeNode rootA, TreeNode rootB){
if(rootA == null && rootB == null){
return true;
}
if(rootA == null || rootB == null){
return false;
}
if(rootA.val != rootB.val){
return false;
}
return isSym(rootA.left,rootB.right) && isSym(rootA.right,rootB.left);
}
}
27. 顺时针打印矩阵
class Solution {
int[] res;
int index = 0;
public int[] spiralOrder(int[][] matrix) {
if(matrix == null || matrix.length <= 0 || matrix[0].length <= 0){
return new int[]{
};
}
int start = 0;
int rows = matrix.length;
int cols = matrix[0].length;
res = new int[rows*cols];
while(rows>start*2 && cols>start*2){
spiralOrderToInt(matrix,cols,rows,start);
start ++;
}
return res;
}
public void spiralOrderToInt(int[][] matrix,int cols,int rows,int start){
int rowEnd = rows-1-start;
int colEnd = cols-1-start;
//1.从左往右打印一行
for(int i=start; i<=colEnd; i++){
res[index] = matrix[start][i];
index ++;
}
//2.从上往下
if(start < rowEnd){
for(int i=start+1; i<=rowEnd; i++){
res[index] = matrix[i][colEnd];
index ++;
}
}
//3.从右往左
if(start<rowEnd && start<colEnd){
for(int i=colEnd-1; i>=start; i--){
res[index] = matrix[rowEnd][i];
index ++;
}
}
//4.从下往上
if(start<colEnd && start+1<rowEnd){
for(int i=rowEnd-1; i>=start+1; i--){
res[index] = matrix[i][start];
index ++;
}
}
}
}
28. 包含min函数的栈
class MinStack {
Stack<Integer> stack;
Stack<Integer> stackAssit;
/** initialize your data structure here. */
public MinStack() {
stack = new Stack();
stackAssit = new Stack();
}
public void push(int x) {
stack.push(x);
if(stackAssit.empty() || stackAssit.peek() > x){
stackAssit.push(x);
} else {
stackAssit.push(stackAssit.peek());
}
}
public void pop() {
if(!stack.empty() && !stackAssit.empty()){
stack.pop();
stackAssit.pop();
}
}
public int top() {
return stack.peek();
}
public int min() {
return stackAssit.peek();
}
}
/**
* Your MinStack object will be instantiated and called as such:
* MinStack obj = new MinStack();
* obj.push(x);
* obj.pop();
* int param_3 = obj.top();
* int param_4 = obj.min();
*/
29. 栈的压入、弹出序列
参考：https://leetcode-cn.com/problems/zhan-de-ya-ru-dan-chu-xu-lie-lcof/solution/mian-shi-ti-31-zhan-de-ya-ru-dan-chu-xu-lie-mo-n-2/
class Solution {
public boolean validateStackSequences(int[] pushed, int[] popped) {
Stack<Integer> stack = new Stack<>();
int i = 0;
for(int num : pushed) {
stack.push(num); // num 入栈
while(!stack.isEmpty() && stack.peek() == popped[i]) {
// 循环判断与出栈
stack.pop();
i++;
}
}
return stack.isEmpty();
}
}
30. 从上到下打印二叉树（1）
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public int[] levelOrder(TreeNode root) {
if(root == null){
return new int[]{
};
}
Queue<TreeNode> queue = new LinkedList<TreeNode>();
queue.offer(root);
List<Integer> resArr = new ArrayList<Integer>();
while(!queue.isEmpty()){
TreeNode node = queue.poll();
resArr.add(node.val);
if(node.left != null){
queue.offer(node.left);
}
if(node.right != null){
queue.offer(node.right);
}
}
int[] res = new int[resArr.size()];
for(int i=0; i<resArr.size(); i++){
res[i] = resArr.get(i);
}
return res;
}
}
31. 从上到下打印二叉树（2）
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public List<List<Integer>> levelOrder(TreeNode root) {
if(root == null){
List<List<Integer>> list = new ArrayList<>();
return list;
}
Queue<TreeNode> queue = new LinkedList<TreeNode>();
List<Integer> listRow = new ArrayList<Integer>();
List<List<Integer>> list = new ArrayList<>();
int numNowLine = 1;
int numAfterLine = 0;
queue.offer(root);
while(!queue.isEmpty()){
TreeNode node = queue.poll();
listRow.add(node.val);
numNowLine --;
if(node.left != null){
queue.offer(node.left);
numAfterLine ++;
}
if(node.right != null){
queue.offer(node.right);
numAfterLine ++;
}
if(numNowLine == 0){
list.add(listRow);
listRow = new ArrayList<Integer>();
numNowLine = numAfterLine;
numAfterLine = 0;
}
}
return list;
}
}
32. 从上到下打印二叉树（3）之字形
用了两个栈来回的方法
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public List<List<Integer>> levelOrder(TreeNode root) {
if(root == null){
List<List<Integer>> list = new ArrayList<>();
return list;
}
Stack<TreeNode> stack1 = new Stack<TreeNode>();
Stack<TreeNode> stack2 = new Stack<TreeNode>();
List<List<Integer>> list = new ArrayList<>();
stack1.push(root);
//int lines = 1;
while(!stack1.empty() || !stack2.empty()){
List<Integer> listRow = new ArrayList<Integer>();
TreeNode node;
while(!stack1.empty()){
node = stack1.pop();
listRow.add(node.val);
if(node.left != null){
stack2.push(node.left);
}
if(node.right != null){
stack2.push(node.right);
}
}
if(listRow.size() > 0){
list.add(listRow);
}
listRow = new ArrayList<Integer>();
while(!stack2.empty()){
node = stack2.pop();
listRow.add(node.val);
if(node.right != null){
stack1.push(node.right);
}
if(node.left != null){
stack1.push(node.left);
}
}
if(listRow.size() > 0){
list.add(listRow);
}
}
return list;
}
}
33. 二叉搜索树的后序遍历序列
class Solution {
public boolean verifyPostorder(int[] postorder) {
return verifySub(postorder, 0, postorder.length-1);
}
public boolean verifySub(int[] postorder, int i, int j){
if(i >= j) return true;
int p = i;
while(postorder[p] < postorder[j]) p++;
int m = p;
while(postorder[p] > postorder[j]) p++;
return p == j && verifySub(postorder, i, m - 1) && verifySub(postorder, m, j - 1);
}
}
34. 二叉树中和为某一值的路径
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
/*
class Solution {
List> list = new LinkedList<>();
public List> pathSum(TreeNode root, int sum) {
if(root == null){
return list;
}
LinkedList path = new LinkedList();
int currentSum = 0;
findPath(root,sum,path,currentSum);
return list;
}
public void findPath(TreeNode root, int sum, LinkedList path, int currentSum){
currentSum = currentSum+root.val;
path.add(root.val);
boolean isLeaf = root.left==null && root.right==null;
if(currentSum==sum && isLeaf){
list.add(path);
path = new LinkedList();
}
if(root.left != null){
findPath(root.left,sum,path,currentSum);
}
if(root.right != null){
findPath(root.right,sum,path,currentSum);
}
path.removeLast();
}
}*/
class Solution {
LinkedList<List<Integer>> res = new LinkedList<>();
LinkedList<Integer> path = new LinkedList<>();
public List<List<Integer>> pathSum(TreeNode root, int sum) {
recur(root, sum);
return res;
}
void recur(TreeNode root, int tar) {
if(root == null) return;
path.add(root.val);
tar -= root.val;
if(tar == 0 && root.left == null && root.right == null)
res.add(new LinkedList(path));
recur(root.left, tar);
recur(root.right, tar);
path.removeLast();
}
}
35. 复杂链表的复制
/*
// Definition for a Node.
class Node {
int val;
Node next;
Node random;
public Node(int val) {
this.val = val;
this.next = null;
this.random = null;
}
}
*/
/*
// Definition for a Node.
class Node {
int val;
Node next;
Node random;
public Node(int val) {
this.val = val;
this.next = null;
this.random = null;
}
}
*/
class Solution {
public Node copyRandomList(Node head) {
if(head==null){
return null;
}
copy(head);
randomDirect(head);
return reList(head);
}
//拷贝链表
private void copy(Node head){
while(head!=null){
Node cloneNode = new Node(head.val);
Node nextNode = head.next;
head.next = cloneNode;
cloneNode.next = nextNode;
head = cloneNode.next;
}
}
//指定随机指针
private void randomDirect(Node head){
while(head!=null){
Node cloneNode = head.next;
if(head.random!=null){
Node direct = head.random;
cloneNode.random = direct.next;
}
head = cloneNode.next;
}
}
//重新连接 链表
private Node reList(Node head){
Node cloneNode = head.next;
Node cloneHead = cloneNode;
head.next = cloneNode.next;
head = head.next;
while(head!=null){
cloneNode.next = head.next;
head.next = head.next.next;
head = head.next;
cloneNode = cloneNode.next;
}
return cloneHead;
}
}
36. 二叉搜索树与双向链表
/*
// Definition for a Node.
class Node {
public int val;
public Node left;
public Node right;
public Node() {}
public Node(int _val) {
val = _val;
}
public Node(int _val,Node _left,Node _right) {
val = _val;
left = _left;
right = _right;
}
};
*/
class Solution {
Node head, pre;
public Node treeToDoublyList(Node root) {
if(root==null) return null;
dfs(root);
pre.right = head;
head.left =pre;//进行头节点和尾节点的相互指向，这两句的顺序也是可以颠倒的
return head;
}
public void dfs(Node cur){
if(cur==null) return;
dfs(cur.left);
//pre用于记录双向链表中位于cur左侧的节点，即上一次迭代中的cur,当pre==null时，cur左侧没有节点,即此时cur为双向链表中的头节点
if(pre==null) head = cur;
//反之，pre!=null时，cur左侧存在节点pre，需要进行pre.right=cur的操作。
else pre.right = cur;
cur.left = pre;//pre是否为null对这句没有影响,且这句放在上面两句if else之前也是可以的。
pre = cur;//pre指向当前的cur
dfs(cur.right);//全部迭代完成后，pre指向双向链表中的尾节点
}
}
37. 序列化/反序列化二叉树
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Codec {
public String serialize(TreeNode root) {
if(root == null) return "[]";
StringBuilder res = new StringBuilder("[");
Queue<TreeNode> queue = new LinkedList<>() {
{
add(root); }};
while(!queue.isEmpty()) {
TreeNode node = queue.poll();
if(node != null) {
res.append(node.val + ",");
queue.add(node.left);
queue.add(node.right);
}
else res.append("null,");
}
res.deleteCharAt(res.length() - 1);
res.append("]");
return res.toString();
}
public TreeNode deserialize(String data) {
if(data.equals("[]")) return null;
String[] vals = data.substring(1, data.length() - 1).split(",");
TreeNode root = new TreeNode(Integer.parseInt(vals[0]));
Queue<TreeNode> queue = new LinkedList<>() {
{
add(root); }};
int i = 1;
while(!queue.isEmpty()) {
TreeNode node = queue.poll();
if(!vals[i].equals("null")) {
node.left = new TreeNode(Integer.parseInt(vals[i]));
queue.add(node.left);
}
i++;
if(!vals[i].equals("null")) {
node.right = new TreeNode(Integer.parseInt(vals[i]));
queue.add(node.right);
}
i++;
}
return root;
}
}
// Your Codec object will be instantiated and called as such:
// Codec codec = new Codec();
// codec.deserialize(codec.serialize(root));
38. 字符串的排列
class Solution {
//为了让递归函数添加结果方便，定义到函数之外，这样无需带到递归函数的参数列表中
List<String> list = new ArrayList<>();
//同；但是其赋值依赖c，定义声明分开
char[] c;
public String[] permutation(String s) {
c = s.toCharArray();
//从第一层开始递归
dfs(0);
//将字符串数组ArrayList转化为String类型数组
return list.toArray(new String[list.size()]);
}
private void dfs(int x) {
//当递归函数到达第三层，就返回，因为此时第二第三个位置已经发生了交换
if (x == c.length - 1) {
//将字符数组转换为字符串
list.add(String.valueOf(c));
return;
}
//为了防止同一层递归出现重复元素
HashSet<Character> set = new HashSet<>();
//这里就很巧妙了,第一层可以是a,b,c那么就有三种情况，这里i = x,正巧dfs(0)，正好i = 0开始
// 当第二层只有两种情况，dfs(1）i = 1开始
for (int i = x; i < c.length; i++){
//发生剪枝，当包含这个元素的时候，直接跳过
if (set.contains(c[i])){
continue;
}
set.add(c[i]);
//交换元素，这里很是巧妙，当在第二层dfs(1),x = 1,那么i = 1或者 2， 不是交换1和1，要就是交换1和2
swap(i,x);
//进入下一层递归
dfs(x + 1);
//返回时交换回来，这样保证到达第1层的时候，一直都是abc。这里捋顺一下，开始一直都是abc，那么第一位置总共就3个交换
//分别是a与a交换，这个就相当于 x = 0, i = 0;
// a与b交换 x = 0, i = 1;
// a与c交换 x = 0, i = 2;
//就相当于上图中开始的三条路径
//第一个元素固定后，每个引出两条路径,
// b与b交换 x = 1, i = 1;
// b与c交换 x = 1, i = 2;
//所以，结合上图，在每条路径上标注上i的值，就会非常容易好理解了
swap(i,x);
}
}
private void swap(int i, int x) {
char temp = c[i];
c[i] = c[x];
c[x] = temp;
}
}
39. 数组中出现次数超过一半的数字
class Solution {
public int majorityElement(int[] nums) {
int cnt = 1;
int element = nums[0];
for(int i=1; i<nums.length; i++){
if(cnt == 0){
element = nums[i];
cnt = 1;
}
else if(nums[i] == element){
cnt ++;
} else {
cnt --;
}
}
//还可以验证一下这个是不是大于一半多的数字
return element;
}
}
40. 最小的k个数
class Solution {
/*
//用快速选择排序来实现
public int[] getLeastNumbers(int[] arr, int k) {
int[] res = new int[k];
if(arr.length <= k){
return arr;
}
if(k == 0){
return res;
}
for(int i=0; i arr[j]){
min = arr[j];
minIndex = j;
}
}
if(minIndex != i){
arr[minIndex] = arr[i];
arr[i] = min;
}
}
for(int i=0; i
//用一个队列来维护
public int[] getLeastNumbers(int[] arr, int k) {
int[] vec = new int[k];
if (k == 0) {
// 排除 0 的情况
return vec;
}
PriorityQueue<Integer> queue = new PriorityQueue<Integer>(new Comparator<Integer>() {
public int compare(Integer num1, Integer num2) {
return num2 - num1;
}
});
for (int i = 0; i < k; ++i) {
queue.offer(arr[i]);
}
for (int i = k; i < arr.length; ++i) {
if (queue.peek() > arr[i]) {
queue.poll();
queue.offer(arr[i]);
}
}
for (int i = 0; i < k; ++i) {
vec[i] = queue.poll();
}
return vec;
}
}
41. 数据流中的中位数
参考：https://leetcode-cn.com/problems/shu-ju-liu-zhong-de-zhong-wei-shu-lcof/solution/mian-shi-ti-41-shu-ju-liu-zhong-de-zhong-wei-shu-y/
class MedianFinder {
Queue<Integer> A, B;
public MedianFinder() {
A = new PriorityQueue<>(); // 小顶堆，保存较大的一半
B = new PriorityQueue<>((x, y) -> (y - x)); // 大顶堆，保存较小的一半
}
public void addNum(int num) {
if(A.size() != B.size()) {
A.add(num);
B.add(A.poll());
} else {
B.add(num);
A.add(B.poll());
}
}
public double findMedian() {
return A.size() != B.size() ? A.peek() : (A.peek() + B.peek()) / 2.0;
}
}
/**
* Your MedianFinder object will be instantiated and called as such:
* MedianFinder obj = new MedianFinder();
* obj.addNum(num);
* double param_2 = obj.findMedian();
*/
42. 连续子数组的最大和
动态规划
class Solution {
public int maxSubArray(int[] nums) {
int res = nums[0];
for(int i=1; i<nums.length; i++){
if(nums[i-1] > 0){
nums[i] = nums[i-1] +nums[i];
}
if(res < nums[i]){
res = nums[i];
}
}
return res;
}
}
43. 1-n整数中1出现的次数
参考讲解：https://leetcode-cn.com/problems/1nzheng-shu-zhong-1chu-xian-de-ci-shu-lcof/solution/mian-shi-ti-43-1n-zheng-shu-zhong-1-chu-xian-de-2/
class Solution {
public int countDigitOne(int n) {
int res = 0;
int cur = n%10;
int high = n/10;
int low = 0;
int digit = 1;
while(high!=0 || low!=n){
if(cur == 0){
res += high*digit;
} else if(cur == 1){
res += high*digit + low + 1;
} else{
res += (high+1)*digit;
}
low += cur*digit;
cur = high%10;
high = high/10;
digit *= 10;
}
return res;
}
}
44. 数字序列中某一位的数字
class Solution {
public int findNthDigit(int n) {
if(n == 0){
return 0;
}
int digit = 1;
long start = 1;
long count = 9;
while(n > count){
n -= count;
digit ++;
start *= 10;
count = 9*start*digit;
}
long num = start+(n-1)/digit;
return Long.toString(num).charAt((n-1)%digit)-'0';
}
}
45. 把数组排成最小的数
/*
class Solution {
public String minNumber(int[] nums) {
String[] strs = new String[nums.length];
for(int i = 0; i < nums.length; i++)
strs[i] = String.valueOf(nums[i]);
fastSort(strs, 0, strs.length - 1);
StringBuilder res = new StringBuilder();
for(String s : strs)
res.append(s);
return res.toString();
}
void fastSort(String[] strs, int l, int r) {
if(l >= r) return;
int i = l, j = r;
String tmp = strs[i];
while(i < j) {
while((strs[j] + strs[l]).compareTo(strs[l] + strs[j]) >= 0 && i < j) j--;
while((strs[i] + strs[l]).compareTo(strs[l] + strs[i]) <= 0 && i < j) i++;
tmp = strs[i];
strs[i] = strs[j];
strs[j] = tmp;
}
strs[i] = strs[l];
strs[l] = tmp;
fastSort(strs, l, i - 1);
fastSort(strs, i + 1, r);
}
}
*/
class Solution {
public String minNumber(int[] nums) {
String[] strs = new String[nums.length];
for(int i = 0; i < nums.length; i++)
strs[i] = String.valueOf(nums[i]);
Arrays.sort(strs, (x, y) -> (x + y).compareTo(y + x));
StringBuilder res = new StringBuilder();
for(String s : strs)
res.append(s);
return res.toString();
}
}
46. 把数字翻译成字符串
class Solution {
public int translateNum(int num) {
String s = String.valueOf(num);
int dp0 = 1;
int dp1 = 1;
int res = 0;
for(int i=2; i<=s.length(); i++){
String tmp = s.substring(i-2,i);
if(tmp.compareTo("10")>=0 && tmp.compareTo("25")<=0){
res = dp0 + dp1;
} else{
res = dp1;
}
dp0 = dp1;
dp1 = res;
}
return dp1;
}
}
47. 礼物的最大价值
class Solution {
public int maxValue(int[][] grid) {
int m = grid.length;
int n = grid[0].length;
int[][] dp = new int[m][n];
for(int i=0; i<m; i++){
for(int j=0; j<n; j++){
int up = 0;
int left = 0;
if(i > 0){
up = dp[i-1][j];
}
if(j > 0){
left = dp[i][j-1];
}
dp[i][j] = Math.max(up,left) + grid[i][j];
}
}
return dp[m-1][n-1];
}
}
48. 最长不含重复字符的子字符串
class Solution {
public int lengthOfLongestSubstring(String s) {
Map<Character, Integer> dic = new HashMap<>();
int res = 0, tmp = 0;
for(int j = 0; j < s.length(); j++) {
int i = dic.getOrDefault(s.charAt(j), -1); // 获取索引 i
dic.put(s.charAt(j), j); // 更新哈希表
tmp = tmp < j - i ? tmp + 1 : j - i; // dp[j - 1] -> dp[j]
res = Math.max(res, tmp); // max(dp[j - 1], dp[j])
}
return res;
}
}
49. 丑数
class Solution {
public int nthUglyNumber(int n) {
int[] dp = new int[n];
dp[0] = 1;
int a = 0, b = 0, c = 0;
for(int i=1; i<n; i++){
int n2=dp[a]*2, n3=dp[b]*3, n5=dp[c]*5;
dp[i] = Math.min(Math.min(n2,n3),n5);
while(dp[a]*2 <= dp[i]){
a++;
}
while(dp[b]*3 <= dp[i]){
b++;
}
while(dp[c]*5 <= dp[i]){
c++;
}
}
return dp[n-1];
}
}
50. 第一个只出现一次的字符
/*写法一
class Solution {
public char firstUniqChar(String s) {
if(s.length() == 0 || s == null){
return ' ';
}
char[] arrays = s.toCharArray();
Map map = new HashMap<>();
for(char array: arrays){
if(map.containsKey(array)){
map.put(array,false);
} else {
map.put(array,true);
}
}
for(char array: arrays){
if(map.get(array)){
return array;
}
}
return ' ';
}
}
*/
class Solution {
public char firstUniqChar(String s) {
HashMap<Character, Boolean> dic = new HashMap<>();
char[] sc = s.toCharArray();
for(char c : sc)
dic.put(c, !dic.containsKey(c));
for(char c : sc)
if(dic.get(c)) return c;
return ' ';
}
}
51. 数组中的逆序对
//实际上就是 归并排序
/*有问题
class Solution {
int sum = 0;
public int reversePairs(int[] nums) {
int[] temp = new int[nums.length];
mergeSort(nums,0,nums.length-1,temp);
return sum;
}
//分+合方法
public void mergeSort(int[] arr, int left, int right, int[] temp){
if(left < right){
int mid = (left+right)/2; //中间索引
//向左递归进行分解
mergeSort(arr,left,mid,temp);
//向右递归进行分解
mergeSort(arr,mid+1,right,temp);
//合并
merge(arr,left,mid,right,temp);
}
}
//合并的方法, 从后往前比
public void merge(int[] arr, int left, int mid, int right, int[] temp){
int i = mid; //初始化i，左边有序序列的初始索引
int j = right; //初始化j,右边有序序列的初始索引
int t = temp.length-1; //指向temp数组的当前索引
//（一）
//先把左右两边（有序）的数据按照规则填充到temp数组
//直到左右两边的有序序列，有一边处理完毕为止
while(i>=0 && j>mid){
if(arr[i] >= arr[j]){
temp[t] = arr[i];
t --;
i --;
sum += j-mid;
}else{
temp[t] = arr[j];
t --;
j --;
}
}
//（二）
//把有剩余的一边的数据依次全部填充到temp
while(i >=0){
temp[t] = arr[i];
t --;
i --;
}
while(j > mid){
temp[t] = arr[j];
t --;
j --;
}
//（三）
//将temp数组的元素拷贝到arr
t = 0;
int tempLeft = left;
while(tempLeft <= right){
arr[tempLeft] = temp[t];
t ++;
tempLeft ++;
}
}
}
*/
//实际上就是 归并排序的思想
//照抄剑指offer代码
class Solution {
public int reversePairs(int[] nums) {
if(nums.length < 2){
return 0;
}
int[] temp = new int[nums.length];
for(int i=0; i<nums.length; i++){
temp[i] = nums[i];
}
int sum = inversePairsCore(nums,temp,0,nums.length-1);
return sum;
}
public int inversePairsCore(int[] nums, int[] temp, int start,int end){
if(start == end){
temp[start] = nums[start];
return 0;
}
int length = (end - start)/2;
int left = inversePairsCore(temp,nums,start,start+length);
int right = inversePairsCore(temp,nums,start+length+1,end);
int i = start+length;
int j = end;
int indexCopy = end;
int count = 0;
while(i >= start && j >= start+length+1){
if(nums[i] > nums[j]){
temp[indexCopy--] = nums[i--];
count += j-start-length;
}else{
temp[indexCopy--] = nums[j--];
}
}
for(;i>=start;--i){
temp[indexCopy--]=nums[i];
}
for(;j>=start+length+1;--j){
temp[indexCopy--]=nums[j];
}
return left+right+count;
}
}
52. 两个链表的第一个公共节点
方法很多：双指针法、先求出两个链表的长度法、栈的方法、集合的方法
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution {
/*先计算长度，再比较
public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
if(headA == null || headB == null){
return null;
}
int cntA = 0;
int cntB = 0;
ListNode tempA = headA;
ListNode tempB = headB;
while(tempA != null){
cntA++;
tempA = tempA.next;
}
while(tempB != null){
cntB++;
tempB = tempB.next;
}
tempA = headA;
tempB = headB;
if(cntA > cntB){
for(int i=0; i
//双指针法
public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
//tempA和tempB我们可以认为是A,B两个指针
ListNode tempA = headA;
ListNode tempB = headB;
while (tempA != tempB) {
//如果指针tempA不为空，tempA就往后移一步。
//如果指针tempA为空，就让指针tempA指向headB（注意这里是headB不是tempB）
tempA = tempA == null ? headB : tempA.next;
//指针tempB同上
tempB = tempB == null ? headA : tempB.next;
}
//tempA要么是空，要么是两链表的交点
return tempA;
}
}
53. 在排序数组中查找数字（1）
先找出右端点，再找出左端点
class Solution {
public int search(int[] nums, int target) {
int i = 0;
int j = nums.length-1;
int left = 0;
int right = 0;
while(i <= j){
int mid = (i+j)/2;
if(nums[mid] <= target){
i = mid+1;
}else{
j = mid-1;
}
}
right = i;
// 若数组中无 target ，则提前返回
if(j >= 0 && nums[j] != target) return 0;
i = 0;
while(i <= j){
int mid = (i+j)/2;
if(nums[mid] < target){
i = mid+1;
}else{
j = mid-1;
}
}
left = j;
return right-left-1;
}
}
54. 在排序数组中查找数字（2）
class Solution {
public int missingNumber(int[] nums) {
int i = 0, j = nums.length-1;
while(i <= j){
int m = (i+j)/2;
if(nums[m] == m){
//注意这里的m==j和下面的m==0的边界条件
if(m == j || nums[m+1] != (m+1)){
return m+1;
}
i = m+1;
}else{
if(m == 0 || nums[m-1] == (m-1)){
return m;
}
j = m-1;
}
}
return i;
}
}
55. 二叉搜索树的第k大节点
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
/*我的写法
class Solution {
List list = new ArrayList();
public int kthLargest(TreeNode root, int k) {
mid(root);
return list.get(k-1).val;
}
public void mid(TreeNode root){
if(root == null){
return;
}
mid(root.right);
list.add(root);
mid(root.left);
}
}
*/
//大佬的写法，可以提前终止
class Solution {
int res, k;
public int kthLargest(TreeNode root, int k) {
this.k = k;
dfs(root);
return res;
}
void dfs(TreeNode root) {
if(root == null) return;
dfs(root.right);
if(k == 0) return;
if(--k == 0) res = root.val;
dfs(root.left);
}
}
56. 二叉树的深度
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
/*
//深度有限遍历
class Solution {
public int maxDepth(TreeNode root) {
if(root == null) return 0;
return Math.max(maxDepth(root.left), maxDepth(root.right)) + 1;
}
}
*/
//层序遍历
class Solution {
public int maxDepth(TreeNode root) {
if(root == null) return 0;
List<TreeNode> queue = new LinkedList<>() {
{
add(root); }}, tmp;
int res = 0;
while(!queue.isEmpty()) {
tmp = new LinkedList<>();
for(TreeNode node : queue) {
if(node.left != null) tmp.add(node.left);
if(node.right != null) tmp.add(node.right);
}
queue = tmp;
res++;
}
return res;
}
}
57. 平衡二叉树
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public boolean isBalanced(TreeNode root) {
return recur(root) != -1;
}
private int recur(TreeNode root) {
if (root == null) return 0;
int left = recur(root.left);
if(left == -1) return -1;
int right = recur(root.right);
if(right == -1) return -1;
return Math.abs(left - right) < 2 ? Math.max(left, right) + 1 : -1;
}
}
58. 数组中数字出现的次数（1）
class Solution {
public int[] singleNumbers(int[] nums) {
int ret = 0;
for (int n : nums) {
ret ^= n;
}
int div = 1;
while ((div & ret) == 0) {
div <<= 1;
}
int a = 0, b = 0;
for (int n : nums) {
if ((div & n) != 0) {
a ^= n;
} else {
b ^= n;
}
}
return new int[]{
a, b};
}
}
59. 数组中数字出现的次数（2）
class Solution {
public int singleNumber(int[] nums) {
int[] counts = new int[32];
for(int num : nums) {
for(int j = 0; j < 32; j++) {
counts[j] += num & 1;
num >>>= 1;
}
}
int res = 0, m = 3;
for(int i = 0; i < 32; i++) {
res <<= 1;
res |= counts[31 - i] % m;
}
return res;
}
}
60. 和为s的两个数字
class Solution {
public int[] twoSum(int[] nums, int target) {
int i = 0, j = nums.length - 1;
while(i < j) {
int s = nums[i] + nums[j];
if(s < target) i++;
else if(s > target) j--;
else return new int[] {
nums[i], nums[j] };
}
return new int[0];
}
}
61. 和为s的连续正数序列
class Solution {
public int[][] findContinuousSequence(int target) {
List<int[]> vec = new ArrayList<int[]>();
for (int l = 1, r = 2; l < r;) {
int sum = (l + r) * (r - l + 1) / 2;
if (sum == target) {
int[] res = new int[r - l + 1];
for (int i = l; i <= r; ++i) {
res[i - l] = i;
}
vec.add(res);
l++;
} else if (sum < target) {
r++;
} else {
l++;
}
}
return vec.toArray(new int[vec.size()][]);
}
}
62. 翻转单词顺序
/*
//双指针法
class Solution {
public String reverseWords(String s) {
s = s.trim(); // 删除首尾空格
int j = s.length() - 1, i = j;
StringBuilder res = new StringBuilder();
while(i >= 0) {
while(i >= 0 && s.charAt(i) != ' ') i--; // 搜索首个空格
res.append(s.substring(i + 1, j + 1) + " "); // 添加单词
while(i >= 0 && s.charAt(i) == ' ') i--; // 跳过单词间空格
j = i; // j 指向下个单词的尾字符
}
return res.toString().trim(); // 转化为字符串并返回
}
}
*/
//Java内置函数
class Solution {
public String reverseWords(String s) {
String[] strs = s.trim().split(" "); // 删除首尾空格，分割字符串
StringBuilder res = new StringBuilder();
for(int i = strs.length - 1; i >= 0; i--) {
// 倒序遍历单词列表
if(strs[i].equals("")) continue; // 遇到空单词则跳过
res.append(strs[i] + " "); // 将单词拼接至 StringBuilder
}
return res.toString().trim(); // 转化为字符串，删除尾部空格，并返回
}
}
63. 左旋转字符串
/*
//字符串切片
class Solution {
public String reverseLeftWords(String s, int n) {
return s.substring(n, s.length()) + s.substring(0, n);
}
}
*/
//遍历拼接
class Solution {
public String reverseLeftWords(String s, int n) {
StringBuilder res = new StringBuilder();
for(int i = n; i < s.length(); i++)
res.append(s.charAt(i));
for(int i = 0; i < n; i++)
res.append(s.charAt(i));
return res.toString();
}
}
64. 滑动窗口的最大值
class Solution {
public int[] maxSlidingWindow(int[] nums, int k) {
if(nums.length == 0 || k == 0) return new int[0];
Deque<Integer> deque = new LinkedList<>();
int[] res = new int[nums.length - k + 1];
for(int j = 0, i = 1 - k; j < nums.length; i++, j++) {
if(i > 0 && deque.peekFirst() == nums[i - 1])
deque.removeFirst(); // 删除 deque 中对应的 nums[i-1]
while(!deque.isEmpty() && deque.peekLast() < nums[j])
deque.removeLast(); // 保持 deque 递减
deque.addLast(nums[j]);
if(i >= 0)
res[i] = deque.peekFirst(); // 记录窗口最大值
}
return res;
}
}
65. 队列的最大值
class MaxQueue {
Queue<Integer> q;
Deque<Integer> d;
public MaxQueue() {
q = new LinkedList<Integer>();
d = new LinkedList<Integer>();
}
public int max_value() {
if (d.isEmpty()) {
return -1;
}
return d.peekFirst();
}
public void push_back(int value) {
while (!d.isEmpty() && d.peekLast() < value) {
d.pollLast();
}
d.offerLast(value);
q.offer(value);
}
public int pop_front() {
if (q.isEmpty()) {
return -1;
}
int ans = q.poll();
if (ans == d.peekFirst()) {
d.pollFirst();
}
return ans;
}
}
/**
* Your MaxQueue object will be instantiated and called as such:
* MaxQueue obj = new MaxQueue();
* int param_1 = obj.max_value();
* obj.push_back(value);
* int param_3 = obj.pop_front();
*/
66. 扑克牌中的顺子
class Solution {
public boolean isStraight(int[] nums) {
Set<Integer> repeat = new HashSet<>();
int max = 0, min = 14;
for(int num : nums) {
if(num == 0) continue; // 跳过大小王
max = Math.max(max, num); // 最大牌
min = Math.min(min, num); // 最小牌
if(repeat.contains(num)) return false; // 若有重复，提前返回 false
repeat.add(num); // 添加此牌至 Set
}
return max - min < 5; // 最大牌 - 最小牌 < 5 则可构成顺子
}
}
67. 圆圈中最后剩下的数字
/*
//递归
class Solution {
public int lastRemaining(int n, int m) {
return f(n, m);
}
public int f(int n, int m) {
if (n == 1) {
return 0;
}
int x = f(n - 1, m);
return (m + x) % n;
}
}
*/
//迭代
class Solution {
public int lastRemaining(int n, int m) {
int f = 0;
for (int i = 2; i != n + 1; ++i) {
f = (m + f) % i;
}
return f;
}
}
68. 股票的最大利润
class Solution {
public int maxProfit(int[] prices) {
int cost = Integer.MAX_VALUE, profit = 0;
for(int price : prices) {
cost = Math.min(cost, price);
profit = Math.max(profit, price - cost);
}
return profit;
}
}
69. 求1+2+。。。+n
class Solution {
public int sumNums(int n) {
boolean flag = n > 0 && (n += sumNums(n - 1)) > 0;
return n;
}
}
70. 不用加减乘除做加法
class Solution {
public int add(int a, int b) {
while(b != 0) {
// 当进位为 0 时跳出
int c = (a & b) << 1; // c = 进位
a ^= b; // a = 非进位和
b = c; // b = 进位
}
return a;
}
}
71. 构建乘积数组
class Solution {
public int[] constructArr(int[] a) {
if(a.length == 0) return new int[0];
int[] b = new int[a.length];
b[0] = 1;
int tmp = 1;
for(int i = 1; i < a.length; i++) {
b[i] = b[i - 1] * a[i - 1];
}
for(int i = a.length - 2; i >= 0; i--) {
tmp *= a[i + 1];
b[i] *= tmp;
}
return b;
}
}
72. 把字符串转换为整数
class Solution {
public int strToInt(String str) {
char[] c = str.trim().toCharArray();
if(c.length == 0) return 0;
int res = 0, bndry = Integer.MAX_VALUE / 10;
int i = 1, sign = 1;
if(c[0] == '-') sign = -1;
else if(c[0] != '+') i = 0;
for(int j = i; j < c.length; j++) {
if(c[j] < '0' || c[j] > '9') break;
if(res > bndry || res == bndry && c[j] > '7') return sign == 1 ? Integer.MAX_VALUE : Integer.MIN_VALUE;
res = res * 10 + (c[j] - '0');
}
return sign * res;
}
}
73. 二叉搜索树的最近公共祖先
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
while(root != null) {
if(root.val < p.val && root.val < q.val) // p,q 都在 root 的右子树中
root = root.right; // 遍历至右子节点
else if(root.val > p.val && root.val > q.val) // p,q 都在 root 的左子树中
root = root.left; // 遍历至左子节点
else break;
}
return root;
}
}
74. 二叉树的最近公共祖先
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
if(root == null) return null; // 如果树为空，直接返回null
if(root == p || root == q) return root; // 如果 p和q中有等于 root的，那么它们的最近公共祖先即为root（一个节点也可以是它自己的祖先）
TreeNode left = lowestCommonAncestor(root.left, p, q); // 递归遍历左子树，只要在左子树中找到了p或q，则先找到谁就返回谁
TreeNode right = lowestCommonAncestor(root.right, p, q); // 递归遍历右子树，只要在右子树中找到了p或q，则先找到谁就返回谁
if(left == null) return right; // 如果在左子树中 p和 q都找不到，则 p和 q一定都在右子树中，右子树中先遍历到的那个就是最近公共祖先（一个节点也可以是它自己的祖先）
else if(right == null) return left; // 否则，如果 left不为空，在左子树中有找到节点（p或q），这时候要再判断一下右子树中的情况，如果在右子树中，p和q都找不到，则 p和q一定都在左子树中，左子树中先遍历到的那个就是最近公共祖先（一个节点也可以是它自己的祖先）
else return root; //否则，当 left和 right均不为空时，说明 p、q节点分别在 root异侧, 最近公共祖先即为 root
}
}