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//有一种固定的字符串速记方式，如a(b)<3>c表示b重复三次，//即abbbc，可能存在嵌套，如a(b©<3>)<2>d表示abcccbcccd，//输入速记形式，输出原本的字符串。#include <iostream>#include <string>#include<stack>using namespace std;string reverseString(string &s) { //if (s.s

输入dgccc 输出cccdg输入dgggCcc 输出gggccdC输入一个字符串，按照字符串中字符的数量从大到小输出#include<iostream>#include<unordered_map>#include<string>#include<algorithm>using namespace std;bool comp(const pair<char, int>& a, const pair<char,...

#快速排序#include <iostream>#include <vector>using namespace std;int partition(vector<int> &a, int left, int right) { int pivot = a[left]; while (left < right) { while (left<right&&a[right]>pivot) { --right;

层序输入字符串创建二叉树层序输出二叉树示例：输入1 2 3 4 # 5 # # # # # -> 输出[[1][2,3][4,5]]#include<iostream>#include<algorithm>#include<string>#include<queue>using namespace std;class TreeNode{public: int val; TreeNode *left; TreeNode *

给一个二叉树和目标值，判断是否存在从根节点到叶子节点的乘积等于目标值的路径；#include<iostream>#include<vector>#include<algorithm>using namespace std;class TreeNode {public: int val; TreeNode *left; TreeNode *right; TreeNode() :val(0),left(nullptr),right(nullptr){}

反转链表输入只能输入0~9的数字字符串#include<iostream>#include<algorithm>#include<string>#include<queue>using namespace std;class LinkNode{public: int val; LinkNode *next; LinkNode() :val(0), next(nullptr){} LinkNode(int value) :val(val

合并两个有序的数组#include<iostream>#include<algorithm>#include<string>#include<queue>using namespace std;void ABsort(int A[], int B[], int m, int n, int seq) { while (m > 0 && n > 0) { if (A[m - 1] <= B[n - 1]) {

合并两个有序链表#include<iostream>using namespace std;class ListNode {public: int val; ListNode *next; ListNode():val(),next(nullptr){} ListNode(int value):val(value), next(nullptr) {}};ListNode *mergeNode(ListNode *l1, ListNode *l2) { if (!l1&

懒汉模式mutex m;class Singleton{public: static Singleton* getInstance() { if (instance == nullptr) { m.lock(); if (instance == nullptr) { instance = new Singleton(); } m.unlock(); } return instance; }private: Singleton(); ~Single

给定一个二叉树的根节点 root ，和一个整数 targetSum ，求该二叉树里节点值之积等于targetSum 的路径的数目。#include<iostream>#include<vector>#include<algorithm>using namespace std;class TreeNode {public: int val; TreeNode *left; TreeNode *right; TreeNode() :val(0),left

创建链表，判断链表是否有环#include<iostream>#include<algorithm>#include<string>#include<queue>using namespace std;class LinkNode{public: int val; LinkNode *next; LinkNode() :val(0), next(nullptr){} LinkNode(int value) :val(value), ne

字节测开一面五张扑克牌，判断是不是顺子输入：A (1) 2 3 4 5 输出：true输入：10 J(11) Q(12) K(13) A(1) 输出：false#include <iostream>#include<vector>using namespace std;bool islist(vector<char> &a) { for (int i = 1; i < 5; i++) { if (a[i]-a[i - 1] == 1)

输入输出#include<iostream>#include<vector>#include<algorithm>using namespace std;int main() { int m, n; vector<int> res; while (cin >> m) { res.emplace_back(m); if (cin.get() == '\n') { break; } } cin &gt

k个一组链表反转#include<iostream>#include<algorithm>#include<string>#include<queue>using namespace std;class LinkNode{public: int val; LinkNode *next; LinkNode() :val(0), next(nullptr){} LinkNode(int value) :val(value), next(nu

给一个字符串，提取出其中偶数位的字符，组成一个新的字符串并打印出来，另外要检验输入字符串的合法性，即是否只包含数字、大小写字母，且长度小于20，如果是非法字符串，打印ERROR。#include<iostream>#include<algorithm>#include<string>#include<queue>using namespace std;int main() { string input; getline(cin,input

给一个二叉树和目标值，找出所有从根节点到叶子节点路径总积等于给定目标和的路径。#include<iostream>#include<vector>#include<algorithm>using namespace std;class TreeNode {public: int val; TreeNode *left; TreeNode *right; TreeNode() :val(0),left(nullptr),right(nullptr){}

实现多线程同步1普通加锁mutex m1;void thread1(int a) { for (int i = 0; i < 100; ++i) { m1.lock(); cout <<"thread1"<<" "<< m++<<endl; m1.unlock(); } }2lock_guard 加锁mutex m1;void thread1(int a) { lock_guard<mutex> l1(

递归生成镜像树#include<iostream>#include<stack>#include<queue>using namespace std;class TreeNode {public: int val; TreeNode *left; TreeNode *right; TreeNode(int value):val(value),left(nullptr),right(nullptr){}};TreeNode *mirror(Tre

#include<iostream>#include<vector>using namespace std;int numSubarrayProductLessThanK(vector<int>& nums, int k) { int len = nums.size(); int i = 0; int j = 0; int res = 0; int cur = 1; for (; j < len; j++) { cur *= num

leetcode11. 盛最多水的容器给你 n 个非负整数 a1，a2，…，an，每个数代表坐标中的一个点 (i, ai) 。在坐标内画 n 条垂直线，垂直线 i 的两个端点分别为 (i, ai) 和 (i, 0) 。找出其中的两条线，使得它们与 x 轴共同构成的容器可以容纳最多的水。说明：你不能倾斜容器。输入：[1,8,6,2,5,4,8,3,7]输出：49解释：图中垂直线代表输入数组 [1,8,6,2,5,4,8,3,7]。在此情况下，容器能够容纳水（表示为蓝色部分）的最大值为 49#inc

#堆排序#include <iostream>#include <queue>using namespace std;//堆排序//void adjustHeap(int a[], int parent, int len) {// int temp = a[parent];// int child = 2 * parent + 1;//// while (child <=len ) {// if (child + 1 <= len &&am

#归并排序#include <iostream>using namespace std;void merge(int a[], int temp[], int left, int mid, int right) { int m = right - left + 1; int i = left; int j = mid + 1; int k = 0; while (i <= mid && j <= right) { if (a[i] < a

队列从上到下打印二叉树//bfs打印树节点#include#include#includeusing namespace std;class TreeNode {public:int val;TreeNode *left;TreeNode *right;TreeNode(int value):val(value),left(nullptr),right(nullptr){}};void bfs(TreeNode root, vector<vector> &res