ccbka的博客

#include using namespace std;class Solution {public:    int singleNumber(vector& A) {        int result=A[0];        for (int i=0;i        {            result=result^A[i];        }

#includeusing namespace std;int zdgys(int m, int n){if ((m%n) == 0){return n;}else{int re=zdgys(n, m%n);return re;}}int main(){int m, n;cin >> m >> n;cout return

#includeusing  namespace std;typedef struct node{int data;struct node* left;struct node* right;}Node;typedef struct tree{Node* root=NULL;}Tree;void insert(Tree* tree,int value)

#includeusing  namespace std;typedef struct node{int data;struct node* left;struct node* right;}Node;typedef struct tree{Node* root=NULL;}Tree;void insert(Tree* tree,int value)

#includeusing namespace std;typedef struct node{int data;node* left;node* right;}Node;void preorder(Node* root){if (root != NULL){cout data preorder(root->left);preorder(ro

#include#includeusing namespace std;int main(){int n;typedef struct node{int data;node* left;node* right;}Node;Node n1;cout coutcout return 0;}

#include#includeusing namespace std;class solution{public:int binary_find(vector v,int key,int start,int end){if (start {int mid =(end +start) / 2;if (key == v[mid])return mid;

#include#includeusing namespace std;class solution {public:void heaping(vector& v, int i,int size)//在假设的基础上构建i节点为根节点的最大堆{if (i{int left = 2 * i + 1;int right = 2 * i + 2;int la

要求：在矩形货架上取货，货物重量随机，小车只能向比之前货物更轻的方向行走，求小车能走的最长路径。#include#include#include#include#include#includeusing namespace std;class solution{public:bool pathcore(int* cargo, int i, int j, in

#include#includeusing namespace std;void bubble_sort(vector& v, int L, int R){if (L == R){return;}for (int i = L; i {if (v[i] > v[i + 1]){int temp;temp = v[i + 1];v[i + 1

#include#include#include#include#include#includeusing namespace std;class solution{public:bool check(string s){unordered_map> map_a;for (int i = 0; i {map_a[s[i]].push_ba

#include   #include   #include   using namespace std;    int main() {      vectorint> a;      vectorint> b;      int n;      int tmp;      int i;      cin >> n;        for (i

#include#include#include#includeusing namespace std;class solution{public:bool num_ser( int nums[],int length){map cha; sort(nums,nums+length);for (int i = 0; i {int a = nu

如果一个数列S满足对于所有的合法的i,都有S[i + 1] = S[i] + d, 这里的d也可以是负数和零,我们就称数列S为等差数列。小易现在有一个长度为n的数列x,小易想把x变为一个等差数列。小易允许在数列上做交换任意两个位置的数值的操作,并且交换操作允许交换多次。但是有些数列通过交换还是不能变成等差数列,小易需要判别一个数列是否能通过交换操作变成等差数列#include#in

小易有一些彩色的砖块。每种颜色由一个大写字母表示。各个颜色砖块看起来都完全一样。现在有一个给定的字符串s,s中每个字符代表小易的某个砖块的颜色。小易想把他所有的砖块排成一行。如果最多存在一对不同颜色的相邻砖块,那么这行砖块就很漂亮的。请你帮助小易计算有多少种方式将他所有砖块排成漂亮的一行。(如果两种方式所对应的砖块颜色序列是相同的,那么认为这两种方式是一样的。)例如: s = "ABAB",那

#include#includeusing namespace std;class solution{public:void bubble_sort(vector& arr,int R){if (R>0){for (int i = 0; i {if (arr[i] > arr[i + 1]){int temp;temp = arr[i];

#include#includeusing namespace std;class solution{public:vector bubble_sort(vector& arr){int arr_len=arr.size();for (int i = 0; i {for (int j = 0; j{if (arr[j]>arr[j + 1])

#include#includeusing namespace std;class solution{public:void quick_sort(vector& nums,int L,int R){int M = (L + R) / 2;int i = L;int j = R;int temp = nums[M];while (i{whil

#include#include#include#includeusing namespace std;int main(){int s;//数字转字符串vector v = { 1, 2, 3 };stringstream ss;for (int i = 0; i {ss }string sss;sss = ss.str();/

#include#include#include#include#includeusing namespace std;class solution{public:int cdna(string dnas, unordered_map l){int re = 0;//int temp=-1;vector ree(100,0);for (siz

#include#includeusing namespace std;class solution{public:bool hascore(const char* Matrix, char* str, const int cols, const int rows, int row, int col, int& pathlength, bool* visited)//{

#includeusing namespace std;struct Listnode{int m_data;Listnode* m_next;Listnode(int value, Listnode* next = NULL):m_data(value),m_next(next){}};class solution{public:Listnode* Mer

Given a string containing just the characters , determine if the input string is valid.The brackets must close in the correct order, “()” and “()[]{}” are all valid but “(]” and “([)]” are not.#

Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.Note: Elements in a triplet (a,b,c) mus

String to Integer (atoi)Dec 27 '11Implement atoi to convert a string to an integer.Hint: Carefully consider all possible input cases. If you want a challenge, please do not see below and a

//最优方法 map查找法,map的查询时间复杂度为LogN#include#include#include#includeusing namespace std;class solution{public:vector twosum(vector&num, int& target){map hash;map::const_iterator iter;

Given an array of integers, return indices of the two numbers such that they add up to a specific target.You may assume that each input would have exactly one solution, and you may not use the sam

Reverse digits of an integer.Example1: x = 123, return 321Example2: x = -123, return -321Note:The input is assumed to be a 32-bit signed integer. Your function should return 0 when the rev

#include#include#includeusing namespace std;struct Binarynode{int m_value;Binarynode* m_left;Binarynode* m_right;Binarynode(int value) :m_value(value), m_left(NULL), m_right(NULL){};

#include#include#include//使用栈using namespace std;template class queue{//模板类的定义public:queue(void);~queue(void);void appendtail(const T& num);//成员函数一般是共有的T deletehead();private:sta

#include#include#include#includeusing namespace std;class solution{public:int max_v(vector stu){vector newl(2*(stu.size()/2),0);size_t num = stu.size();int times = num / 2;bo

#include#include#include#include#includeusing namespace std;class point{public:point(int x1, int y1):x(x1), y(y1){};private:int x;int y;};class solution{public:vector m

#include#include#include#includeusing namespace std;class solution{public:vector min_step(vector x, vector y, int n){vector dist(n, 0);//为向量赋初值vector re(n,INT_MAX);//使用INT_MAX需要添

#include#include#includeusing namespace std;class solution{public:void swap(int data[], int randnum, int end){int temp;temp = data[end];data[end] = data[randnum];data[randnum]=

#include#include#include#includeusing namespace std;templatevoid printpq( const string& str, priorityQ& pr){cout cout cout while (!pr.empty()){cout pr.pop();}cout }

网易笔试：堆棋子题目描述小易将n个棋子摆放在一张无限大的棋盘上。第i个棋子放在第x[i]行y[i]列。同一个格子允许放置多个棋子。每一次操作小易可以把一个棋子拿起并将其移动到原格子的上、下、左、右的任意一个格子中。小易想知道要让棋盘上出现有一个格子中至少有i(1 ≤ i ≤ n)个棋子所需要的最少操作次数.输入描述: 输入包括三行,第一行一个整数n(1 ≤ n ≤

#include#include#include#includeusing namespace std;class solution{public:int randominrange(int a, int b){srand(NULL);//随机数的产生，需要种子int num = a + rand() % (b - a + 1);return num;

小易有一个长度为n的整数序列,a_1,...,a_n。然后考虑在一个空序列b上进行n次以下操作:1、将a_i放入b序列的末尾2、逆置b序列小易需要你计算输出操作n次之后的b序列。 输入描述:输入包括两行,第一行包括一个整数n(2 ≤ n ≤ 2*10^5),即序列的长度。第二行包括n个整数a_i(1 ≤ a_i ≤ 10^9),即序列a中的每个整数,以空格分割。