力扣题目汇总
文章目录
- 1315. Sum of Nodes with Even-Valued Grandparent
- DFS
- 473. 火柴拼正方形
- 494. 目标和
- 46. Permutations
- 北京大学 OJ:
- 链表
- 反转链表
- 相交链表
- 5. 最长回文子串
1315. Sum of Nodes with Even-Valued Grandparent
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
int sumEvenGrandparent(struct TreeNode* root){
return getSum(root);
}
int getSum(struct TreeNode* root) {
int sum = 0;
if(root==NULL) {
return sum;
}
struct TreeNode* left = root->left;
struct TreeNode* right = root->right;
if (root->val%2 == 0) {
if (left != NULL) {
struct TreeNode* left_left = left->left;
struct TreeNode* left_right = left->right;
if (left_left != NULL) {
sum += left_left->val;
}
if (left_right != NULL) {
sum += left_right->val;
}
}
if (right != NULL) {
struct TreeNode* right_left = right->left;
struct TreeNode* right_right = right->right;
if (right_left != NULL) {
sum += right_left->val;
}
if (right_right != NULL) {
sum += right_right->val;
}
}
}
sum += getSum(root->left) + getSum(root->right);
return sum;
}
DFS
473. 火柴拼正方形
473. 火柴拼正方形
int edge_num = 0; // 已组成边的个数
int edge_len = 0; // 边的长度
bool dfs(int idx, int curr_len, int *nums, int *vis, int numsSize);
int cmpfunc (const void * a, const void * b)
{
return ( *(int*)a - *(int*)b );
}
bool makesquare(int* nums, int numsSize){
int i;
int sum = 0;
qsort(nums, numsSize, sizeof(int), cmpfunc);
for (i=0; i<numsSize; i ++) {
sum += nums[i];
}
if (sum%4 != 0) {
return false;
}
int *vis = (int *)malloc(numsSize*sizeof(int));
memset(vis, 0, numsSize*sizeof(int));
edge_len = sum/4;
edge_num = 0;
return dfs(0, 0, nums, vis, numsSize);
}
/*
* idx: 从该索引开始搜索。
* curr_len：当前组成的长度。
* vis: 记录某个索引数值是否使用过。
* numsSize：数组大小。
*/
bool dfs(int idx, int curr_len, int *nums, int *vis, int numsSize) {
if (edge_num == 4) {
// 若组成边长为4， 则完成搜索。
return true;
}
if (edge_num < 4) {
for (int i=idx; i<numsSize; i ++) {
if (vis[i] == 1) {
continue;
}
if (i>0 && vis[i-1]==0 && nums[i-1]==nums[i]) {
continue;
}
if (nums[i]+curr_len < edge_len) {
vis[i] = 1;
bool flag = dfs(i+1, nums[i]+curr_len, nums, vis, numsSize);
if (flag == false) {
//将索引为i的数值加入后，其他的数据无法组成边长，则将索引为i的数值退出，进行回溯处理
vis[i] = 0;
} else {
return true;
}
}
if (nums[i]+curr_len == edge_len) {
vis[i] = 1;
edge_num ++;
bool flag = dfs(0, 0, nums, vis, numsSize);
if (flag == true) {
return true;
}
// 回溯
vis[i] = 0;
edge_num --;
}
}
}
return false;
}
494. 目标和
494. 目标和
int sumways = 0;
int cmp(const void *a, const void *b) {
return (*(int *)b) - (*(int *)a);
}
int findTargetSumWays(int* nums, int numsSize, int S){
int i = 0;
qsort(nums, numsSize, sizeof(int), cmp);
int *sum = (int *)malloc(numsSize * sizeof(int));
for (i=numsSize-1; i>=0; i --) {
if (i == numsSize-1) {
sum[i] = nums[i];
continue;
}
sum[i] = sum[i+1] + nums[i];
}
sumways = 0;
dfs(0, 0, nums, numsSize, S, sum);
return sumways;
}
void dfs(int idx, int curr_value, int *nums, int numsSize, int S, int *sum) {
if (idx==numsSize-1) {
if (curr_value+nums[idx] == S) {
sumways ++;
}
if (curr_value-nums[idx] == S) {
sumways ++;
}
return ;
}
if (idx < numsSize-1) {
if (curr_value+nums[idx]+sum[idx+1] < S) {
return;
}
if (curr_value-nums[idx]-sum[idx+1] > S) {
return;
}
dfs(idx+1, curr_value+nums[idx], nums, numsSize, S, sum);
dfs(idx+1, curr_value-nums[idx], nums, numsSize, S, sum);
}
return ;
}
46. Permutations
public class Solution {
List<List<Integer>> ls = new ArrayList<>();
public List<List<Integer>> permute(int[] nums) {
int len = nums.length;
for(int i=0; i<len; i ++){
add(nums[i]);
}
return ls;
}
public void add(int n){
int size = ls.size();
if(size==0){
List<Integer> tls = new ArrayList<>();
tls.add(n);
ls.add(tls);
return;
}
List<List<Integer>> lsls = new ArrayList<>();
for(List<Integer> li:ls){
for(int i=0; i<=li.size(); i ++){
List<Integer> tls = new ArrayList<>(li);
tls.add(i, n);
lsls.add(tls);
}
}
ls = lsls;
}
}
北京大学 OJ:
1011-木棒
#include 正方形6*6
#include 链表
反转链表
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
struct ListNode* reverseList(struct ListNode* head){
struct ListNode *p = head;
struct ListNode *pre = NULL;
struct ListNode *ne = NULL;
while (p) {
ne = p->next;
p->next = pre;
pre = p;
p = ne;
}
return pre;
}
相交链表
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
struct ListNode *getIntersectionNode(struct ListNode *headA, struct ListNode *headB) {
if (headA==NULL || headB==NULL) return NULL;
struct ListNode *pa = headA;
struct ListNode *pb = headB;
while (pa && pb) {
pa = pa->next;
pb = pb->next;
}
if (pa == NULL) {
pa = headB;
}
if (pb == NULL) {
pb = headA;
}
while (pa && pb) {
pa = pa->next;
pb = pb->next;
}
if (pa == NULL) {
pa = headB;
}
if (pb == NULL) {
pb = headA;
}
while (pa && pb && pa != pb) {
pa = pa->next;
pb = pb->next;
}
return pa;
}
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
struct ListNode *getIntersectionNode(struct ListNode *headA, struct ListNode *headB) {
if (headA==NULL || headB==NULL) return NULL;
struct ListNode *pa = headA;
struct ListNode *pb = headB;
while (pa != pb) {
pa = pa == NULL ? headB : pa->next;
pb = pb == NULL ? headA : pb->next;
}
return pa;
}
5. 最长回文子串
char * longestPalindrome(char * s){
int len = strlen(s);
int **arr = (int **)malloc(len * sizeof(int *));
for (int i=0; i<len; i ++) {
arr[i] = (int *)malloc(len * sizeof(int));
memset(arr[i], 0, len * sizeof(int));
}
int i1 = 0;
int i2 = 0;
for (int y=0; y<len; y ++) {
for (int x=0; x<=y; x ++) {
if (x == y) {
arr[x][y] = 1;
} else if (x == y-1) {
if (s[x] == s[y]) {
arr[x][y] = 1;
}
} else {
if (x+1<len && y-1>=0 && arr[x+1][y-1] == 1 && s[x] == s[y]) {
arr[x][y] = 1;
}
}
if (arr[x][y] == 1 && y-x > i2-i1) {
i2 = y;
i1 = x;
}
}
}
for (int i=0; i < len; i ++) {
free(arr[i]);
}
free(arr);
arr = NULL;
int resLen = (i2-i1+2);
char *res = (char *)malloc(resLen * sizeof(char));
strncpy(res, s+i1, (resLen-1)*sizeof(char));
res[resLen-1] = '\0';
return res;
}