leetCode解题报告5道题(五)
题目一：
Path Sum
Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.
For example:Given the below binary tree and
sum = 22
,
5
/ \
4 8
/ / \
11 13 4
/ \ \
7 2 1
return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.
AC代码：
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public boolean hasPathSum(TreeNode root, int sum) {
/*如果root == null直接返回false*/
if (root == null){
return false;
}
/*sum 减去当前root的值*/
sum -= root.val;
/*如果已经到了叶子结点，判断sum是否已经减到了0*/
if (root.left == null && root.right == null && sum == 0){
return true;
}
/*flag的话，要判断左右子树是否有路径可以满足，即为
*/
boolean flag = hasPathSum(root.left, sum) || hasPathSum(root.right, sum);
return flag;
}
}Path Sum II
（姐妹题）Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.
For example:Given the below binary tree and
sum = 22
,
5
/ \
4 8
/ / \
11 13 4
/ \ / \
7 2 5 1
return
[ [5,4,11,2], [5,8,4,5] ]
分析：无情递归就OK了，递归结束的条件：当是叶子结点并且sum的值已经变成了0，这时候的List为一组有效的解，加入到结果集合result中。
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
/**
* CSDN: 胖虎 http://blog.csdn.net/ljphhj
*/
public class Solution {
private ArrayList> result = new ArrayList>();
public ArrayList> pathSum(TreeNode root, int sum) {
if (root == null){
return result;
}
calPath(root, sum, new ArrayList());
return result;
}
public void calPath(TreeNode root, int sum, ArrayList list){
sum -= root.val;
list.add(root.val); //当前值加入“可能的解”集合list中
/*如果为叶子结点了，那么就判断是否sum已经等于0了，如果等于0则为一个有效解*/
if (root.left == null && root.right == null && sum == 0){
result.add(list);
return ;
}
/*如果不是叶子结点，则继续往左右子树找*/
if (root.left != null){
calPath(root.left, sum, new ArrayList(list));
}
if (root.right != null){
calPath(root.right, sum, new ArrayList(list));
}
}
} 题目二：
Merge Sorted Array
Given two sorted integer arrays A and B, merge B into A as one sorted array.
Note:
You may assume that A has enough space (size that is greater or equal to m + n) to hold additional elements from B. The number of elements initialized in A and B are m andn respectively.
AC代码：
public class Solution {
public void merge(int A[], int m, int B[], int n) {
if (m == 0){
for (int i=0; i= 0 && pB >= 0){
if (B[pB] > A[pA]){
A[pIndex--] = B[pB--];
}else{
A[pIndex--] = A[pA--];
}
}
/*pA>=0不处理，因为本来就在A数组中*/
/*如果循环结束了，pB>=0则表示B数组还有值没加入到A数组中*/
while (pB >= 0){
A[pIndex--] = B[pB--];
}
}
} 题目三：
Symmetric Tree(对称树)
Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
For example, this binary tree is symmetric:
1 / \ 2 2 / \ / \ 3 4 4 3
But the following is not:
1 / \ 2 2 \ \ 3 3
分析：题目给我们一个二叉树的结构，让我们去判断这个二叉树是否是对称的。这道题目可以有两种做法，递归(424ms)或者用层次遍历树(524ms)的方法
这个题目的递归很容易就看出来了，我就不详细讲，具体看代码里面的注释理解！
（递归）AC代码（424ms）：
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
/*判断左右子树是否对称*/
public boolean checkSymmetric(TreeNode lsubTree,TreeNode rsubTree){
/*几种基本情况的排除*/
if(lsubTree==null && rsubTree==null)
return true;
else if(lsubTree!=null && rsubTree==null)
return false;
else if(lsubTree==null && rsubTree!=null)
return false;
else if(lsubTree.val != rsubTree.val)
return false;
/*再递归看下子树是否对称，只有子树也对称了才可以*/
boolean lt=checkSymmetric(lsubTree.left, rsubTree.right);
boolean rt=checkSymmetric(lsubTree.right, rsubTree.left);
return lt && rt;
}
/*只要左右子树对称那么整个二叉树就对称了*/
public boolean isSymmetric(TreeNode root) {
if(root==null)
return true;
return checkSymmetric(root.left,root.right);
}
}（层次遍历）AC代码(524ms):
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public boolean isSymmetric(TreeNode root) {
Queue queue = new LinkedList();
if (root == null)
return true;
/*加入根结点的左右子结点*/
queue.add(root.left);
queue.add(root.right);
/*层次遍历*/
while (queue.size() != 0){
String s = new String("");
ArrayList list = new ArrayList();
/*把一层的结点都取出来之后放入list集合*/
while (queue.size() != 0){
TreeNode node = queue.remove();
list.add(node);
}
/*把这一层的结点对应的值转换成字符串*/
for (int i=0;i 题目四：
Flatten Binary Tree to Linked List
Given a binary tree, flatten it to a linked list in-place.
For example,
Given
1
/ \
2 5
/ \ \
3 4 6
The flattened tree should look like:
1
\
2
\
3
\
4
\
5
\
6
分析：仔细观察题目给出的例子，不难发现其实这个可以用递归来做，求root结点的flatten tree相当于先求出左子树，再求出右子树，然后将左子树的最后一个结点连接到右子树的第一个结点，将root的右子树连接到左子树的第一个结点上，总而言之这题还是比较简单的，建议做这题的时候可以在纸上画一下图，帮助理解！
情况：
1、root只有左子树
2、root只有右子树
3、root左右子树都有
AC代码：
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public void flatten(TreeNode root) {
transform(root);
return ;
}
public TreeNode transform(TreeNode root){
if (root == null)
return null;
if (root.left == null && root.right == null)
return root;
TreeNode leftChild = null;
TreeNode rightChild = null;
if (root.left != null){
leftChild = transform(root.left);
}
if (root.right != null){
rightChild = transform(root.right);
}
if (leftChild != null){
leftChild.right = root.right;
root.right = root.left;
root.left = null;
}
if (rightChild != null)
return rightChild;
if (leftChild != null)
return leftChild;
return root;
}
}题目五：
Distinct Subsequences
Given a string S and a string T, count the number of distinct subsequences of T in S.
A subsequence of a string is a new string which is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (ie, "ACE" is a subsequence of "ABCDE" while "AEC" is not).
Here is an example:
S = "rabbbit", T = "rabbit"
Return 3.
题意：求出字符串T在字符串S中出现的次数，
S字符串:
0 1 2 3 4 5 6
r a b b b i t
T字符串:
0 1 2 3 4 5
r a b b i t
出现的最大次数是3，三种情况分别为：
S: 0 1 2 3 5 6
S: 0 1 2 4 5 6
S: 0 1 3 4 5 6
这样子，我们很容易知道这其实是一个DP的问题
对应的状态转移方程式
if(S[i] == T[j])
dp[i][j] = dp[i-1][j-1] + dp[i-1][j];
else
dp[i][j] = dp[i-1][j];
AC代码：
public class Solution {
public int numDistinct(String S, String T) {
int n = S.length();
int m = T.length();
if (m > n)
return 0;
int[][] results = new int[n+1][m+1];
for (int i=0; i