liudongshizhang的博客

class Solution {public: vector<int> searchRange(vector<int>& nums, int target) { int left = 0, right = nums.size() - 1; while (left <= right) { int mid = (left + right) / 2; if (target > n.

class Solution {public: string removeKdigits(string num, int k) { if (k == num.size()) return "0"; stack<char>stk; for (int i = 0; i < num.size(); i++) { while (!stk.empty() && stk.top() > num[i.

class Solution {public: void rotate(vector<int>& nums, int k) { k = k % nums.size(); reverse(nums.begin(), nums.begin() + nums.size() - k); reverse(nums.begin() + nums.size() - k, nums.end()); reverse(nums.be.

class Solution {public: int findPeakElement(vector<int>& nums) { int left = 0, right = nums.size() - 1; // 思路：O(logn)一般考虑二分搜索 // 规律一：如果nums[i] > nums[i + 1], 则在 i 之前一定存在峰值元素 // 规律二：如果nums[i] < nums[i - .

class Solution {public: vector<vector<string>>result; vector<string>temp; bool isPalindrome(string substr) { int left = 0, right = substr.size() - 1; while (left < right) { if (substr[left++] !.

class Solution {public: int evalRPN(vector<string>& tokens) { stack<int>arr; for (int i = 0; i < tokens.size(); i++) { if (tokens[i] == "+" || tokens[i] == "-" || tokens[i] == "*" || tokens[i] == "/") {..

/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public: vector<vector<int>&gt.

#include <iostream> #include <queue>#include <thread> #include <mutex> #include <unistd.h> #include <condition_variable> using namespace std;mutex mtx;condition_variable produce

利用完全二叉树的性质，首先判断左子树的最右结点与右子树的最右结点高度，如果相等，只需要插入到左子树即可，否则插入右子树。struct Node { int val; Node *l; Node *r;};Node* insert(Node* root, Node*newnode) { if (!root || !newnode)return NULL; if (!root->l) { root->l = newnode; return root; } else if

一、题意大意就是输入两个整数a和b，输出a/b的循环小数表示以及循环节的长度。二、解题思路关键在于找循环节，用res来存储每一次的商，用mod数组来存储每一次的余数。在计算小数部分时每次都将上一次的余数乘十作为本次的被除数，依次进行下去直到找到某一次的商和余数在前面出现过，即找到了循环节。要注意输出的格式，每个用例最后应该有两个换行符#include<stdio.h>int res[3005], mod[3005];//res数组记录每一次的商，mod数组记录每一次的余数 int

#include <iostream> using namespace std;int main(){ int n; cin >> n; cout << "要得到的菱形" << endl; for (int i = 1; i <= n; ++i){ for (int j = 1; j <= n - i; ++j) cout << " "; for (int j = 1; j <= i; ++j) .

后序遍历解法：/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public: bool isBalanced(Tree..

class Solution {public: vector<vector<string>> solveNQueens(int n) { vector<vector<string>>res; // 存储最终结果 vector<string>temp(n, string(n, '.')); vector<int>col(n, 0); // 存储已有的列值 vec..

class Solution {public: vector<vector<string>> solveNQueens(int n) { vector<vector<string>>res; // 存储最终结果 vector<string>temp(n, string(n, '.')); vector<int>col(n, 0); // 存储已有的列值 vect.

class Solution {public: bool wordBreak(string s, vector<string>& wordDict) { int length = s.size(); // dp[i]表示字符串 s 的前 i 个字符能否拆分成 wordDict vector<bool>dp(length + 1); dp[0] = true; for (int i =..

class Solution {public: int mySqrt(int x) { int left = 0, right = x / 2 + 1; int res = 0; while (left <= right) { int mid = (left + right) >> 1; if (((long long) mid * mid <= x)) { .

/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public: ListNode* reverseKGroup(ListNode* head, int k) { ListNode *.

class Solution {public: int longestPalindromeSubseq(string s) { // dp[i][j]表示s从 i 到 j 代表的字符串的最长回文子序列 int n = s.size(); if (n == 0) return 0; vector<vector<int>>dp(n + 1, vector<int>(n + 1, 0)); .

/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public: bool isPalindrome(ListNode* head) { if (head == NULL) retur.

class Solution {public: vector<int> maxSlidingWindow(vector<int>& nums, int k) { vector<int>maxInWindow; if (nums.size() >= k && k >= 1) { deque<int>indexs; for (int i .

统计因子中5的个数class Solution {public: int trailingZeroes(int n) { int count = 0; while (n >= 5) { count += (n / 5); n /= 5; } return count; }};

class Solution {public: double myPow(double x, long long n) { if (n == 0 ) return 1; if (n < 0) return 1 / myPow(x, -n); // 如果n是奇数 if (n % 2) { return x * myPow(x, n - 1); } return myPow(.

class Solution {public: string minWindow(string s, string t) { if(s.size() < t.size()) return ""; unordered_map<char, int>need; //目标查询的字符以及数量 for (int i = 0; i < t.size(); i++) { need[t[i]]++; }.

class Solution {public: int nthUglyNumber(int n) { if (n < 6) return n; // 搜索起点 int start2 = 1, start3 =1, start5 = 1; vector<int>dp(n + 1, 0); for (int i = 1; i <= 6; i++) { dp[i] = i...

/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public: TreeNode* buildTree(vector<.

/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public: TreeNode* buildTree(vector<.

/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public: int countNodes(TreeNode* root).

/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode() : val(0), next(nullptr) {} * ListNode(int x) : val(x), next(nullptr) {} * ListNode(int x, ListNode *next) : val(x), next(next.

1、对 vector<vector> 排序，需要按照先比较区间开始，如果相同再比较区间结束， 使用默认的排序规则即可2、使用双指针，左边指针指向当前区间的开始3、使用一个变量来记录连续的范围 t4、右指针开始往后寻找，如果后续的区间的开始值比 t 还小，说明重复了，可以归并到一起5、此时更新更大的结束值到 t6、直到区间断开，将 t 作为区间结束，存储到答案里7、然后移动左指针，跳过中间已经合并的区间class Solution {public: vector&l.

class Solution {public: bool canPermutePalindrome(string s) { vector<int>arr(128, 0); for (int i = 0; i < s.size(); i++) { int index = s[i] - 0; arr[index]++; } // 字符个数为奇数的个数 in.

class Solution {public: int ladderLength(string beginWord, string endWord, vector<string>& wordList) { unordered_set<string> wordDict(wordList.begin(), wordList.end()); if (wordDict.find(endWord) == wordDict.end()) ret.

class Solution {public: bool search(vector<int>& nums, int target) { int left = 0; int right = nums.size() - 1; while (left <= right) { int mid = left + (right - left) / 2; if (nums[mid] ==.

class Solution {public: int lengthOfLIS(vector<int>& nums) { if (nums.size() < 1) { return 0; } int max = 1; vector<int>dp(nums.size(), 1); for (int i = 1; i < nums.size(); i++.

class Solution {public: TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) { // 如果根节点为空，则返回空 if (root == NULL) { return NULL; } // 如果根节点是p或者q，则返回root if (root == q || root == p.

/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public: ListNode* mergeKLists(vector<ListNode*>& lists) { .

class Solution {public: int threeSumClosest(vector<int>& nums, int target) { sort(nums.begin(), nums.end()); int minnum = pow(2,31) - 1; int res = 0; for (int i = 0; i < nums.size() - 2; i++) { .

class Solution {public: TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) { int iend = inorder.size() - 1; int postend = postorder.size() - 1; return dfs(inorder, 0, iend, postorder, 0, post.

class Solution {public: TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) { int pend = preorder.size() - 1; int iend = inorder.size() - 1; return dfs(preorder, 0, pend, inorder, 0, iend); .

class Solution {public: vector<vector<string>>result; vector<string>temp_str; bool isPalindrome(string substr) { int left = 0, right = substr.size() - 1; while(left < right) { if (substr[left+.

class Solution {public: void solve(vector<vector<char>>& board) { if(board.size() == 0) return; int row = board.size(); int col = board[0].size(); // 左右边界 for(int i = 0; i < row; i++) { .