cookcoder的博客

class Solution {public: bool wordPattern(string pattern, string str) { int p_len =pattern.size(); vector word ; int t=0; for(int i=0;i<str.size();i++) if (str[i]==' '){ string ww=s

#include#include#include#includeusing namespace std;//先对原始数组进行排序，建立一个新的Id vector存储各个元素排序之前的序号，//对小于target的部分用for循环判断是否为twosum和class Solution {public:    vector twoSum(vector& nums, i

#include#include#include#includeusing namespace std; //从按行的角度出发，计算逐个需要存的字符的对应原串的下标。class Solution {public:Za    string convert(string s, int numRows) {string re= "";int len =s.size

#include#include#include#includeusing namespace std; //从按行的角度出发，计算逐个需要存的字符的对应原串的下标。class Solution {public:    int reverse(int x) {        if(x==0)        return x;        int bit

#include#include#include#includeusing namespace std; //从按行的角度出发，计算逐个需要存的字符的对应原串的下标。class Solution {public: int strStr(string haystack,string needle) { int i,j; // cout << needle.size(

#include#include#include#include#include#includeusing namespace std;class Solution {public:string reverseWords(string str) {if (str == "")return str;else {int i = 0;int j