manguoyu的博客

int lengthOfLongestSubstring(char * s) { int len = strlen(s); if (len <= 1) { return len; } int hash[1024] = {0}; int left = 0; int right = 0; int max = 0; while (s[right] != '\0') { if (hash[s[right]] &.

int getDays(int* weights, int weightsSize, int mid){ int days = 1; int count = 0; for (int i = 0; i < weightsSize; i++) { if ((count + weights[i]) > mid) { days++; count = 0; } count += we.

https://leetcode-cn.com/problems/unique-paths/// 三步走// 1. 确定数组意义：dp[i][j] :从 (0,0) 到 (i,j)有多少条路径// 2. 找出关系式： dp[i][j] = dp[i-1][j] + dp[i][j-1]// 3. 初始值： dp[0][0] = 1// dp[i][0] = 1// dp[0][j] = 1int uniqu

// node就是要删除的节点， 且不是末尾结点// 直接将node->next的值赋给node，同时node->next = node->next->next;void deleteNode(struct ListNode* node) { struct ListNode *next = node->next; node->val = node->next->val; node->next = node->next.

// 记录当前0的个数， 当遍历到非0时， 根据前面有几个0， 既能得知要往前挪几位void moveZeroes(int* nums, int numsSize){ if (NULL == nums || numsSize <= 1) return; int i; int zero_num = 0; for (i = 0; i < numsSize; i++) { if (nums[i] == 0) { zero_n.

利用双指针， 加一个临时最大值。临时最大值 = MIN（height[left] , height[right]）* (right - left), 同时根据哪个矮就往前或者往后挪一位int maxArea(int* height, int heightSize){ if (NULL == height) return 0; if (heightSize < 2) return 0; int left = 0; int right = heightSi.

#define MAX(a,b) (a) > (b) ? (a) : (b)int maxSubArray(int* nums, int numsSize){ int max = nums[0]; int sum = nums[0]; for (int i = 1; i < numsSize; i++) { if (sum < 0) { // 当前sum值小于0, 那么不管怎么加都会把后续的值变小， 所以直接置为0， 重新计算sum值 .

//1. 定义 dp[i]： 为到第i间房最多偷多少//2. 关系： 有两种偷法： 不偷：dp[i] = dp[i-1]// 偷：dp[i] = nums[i] + dp[i-2] // 前面一个房子不能偷// dp[i] = MAX(dp[i-1], nums[i] + dp[i-2])//3. 初始值// 只有一个房间，偷之: dp[0] = num[0]// 只有两个房间，偷其中金额较高的那个房间:.

// 三步走// 1. 确定数组意义：dp[i][j] 当前走到dp[i][j]的最小路径// 2. 找出关系式： dp[i][j] = MIN(dp[i - 1][j], dp[i][j - 1]) + grid[i][j]// 3. 初始值： dp[0][0] = grid[0][0]// dp[i][0] = grid[i][0] + dp[i - 1][0];// dp[0][j] = grid[0][j] .

https://leetcode-cn.com/problems/unique-paths/// 动态规划三步走：// 1. 定义数组a[i][j]: 到a[i][j] 总共有多少条路径// 2. 找出数组元素关系：a[i][j] = a[i-1][j] + a[i][j - 1]// 3. 初始条件: a[0][0] = 0, a[0][j] = 1, a[i][0] = 1int uniquePaths(int m, int n) { if (m == 1 &amp

/** * Return an array of arrays of size *returnSize. * The sizes of the arrays are returned as *returnColumnSizes array. * Note: Both returned array and *columnSizes array must be malloced, assume caller calls free(). */ /*回溯法框架解题：result = []def.

/** * Note: The returned array must be malloced, assume caller calls free(). */ /*回溯法框架解题：result = []def backtrack(路径, 选择列表): if 满足结束条件: result.add(路径) return for 选择 in 选择列表: 做选择 backtrack(路径, 选择列表) .

// 引入min和max// 左子树结点的最大值一定小于root->val// 右子树结点的最小值一定大于root->valbool isValidBSTSub(struct TreeNode *node, int min, int max) { if (NULL == node) return true; if (node->val < min || node->val > max) return false; if (node-&g.

利用栈， 注意边界// 将左括号入栈，一旦遇到右括号就出栈，出栈的元素与左括号不匹配，直接返错bool isValid(char * s){ char stack[10000] = {0}; int top = -1; while (*s != '\0') { if (*s == '(' || *s == '{' || *s == '[') { stack[++top] = *s; } if (*s =

https://leetcode-cn.com/problems/partition-list//** * Definition for singly-linked list. * struct ListNode { * int val; * struct ListNode *next; * }; */struct ListNode* partition(struct ListNode* head, int x){ if (NULL == head) retu

借用临时接节点/** * Definition for singly-linked list. * struct ListNode { * int val; * struct ListNode *next; * }; */struct ListNode* swapPairs(struct ListNode* head){ if (head == NULL || head->next == NULL) return head; struct Li