Minimum Window Substring 最小窗口覆盖所有字串

Minimum Window Substring

Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n).

For example,
S = "ADOBECODEBANC"
T = "ABC"

Minimum window is "BANC".

Note:
If there is no such window in S that covers all characters in T, return the empty string "".

If there are multiple such windows, you are guaranteed that there will always be only one unique minimum window in S.

class Solution {public://双指针滑动 求最小窗口覆盖所有字串//INT_MAX 2147483647 #include<limits.h>    string minWindow(string s, string t) {                int sLen=s.size();        int tLen=t.size();        int tHash[256],dp[256];        memset(tHash,0,sizeof(tHash));//t字符的个数        memset(dp,0,sizeof(dp));//计算字符数        for(int i=0;i<tLen;i++){            tHash[t[i]]++;        }                int count=0;        int min=INT_MAX,start=0;        int left=0,right=0;        for(right=0;right<sLen;right++){            if(tHash[s[right]]==0)//字串没有该字符                continue;            dp[s[right]]++;            if(dp[s[right]]<=tHash[s[right]])//该字符增加一个                count++;                            if(count==tLen){                                while(left<right){                    if(tHash[s[left]]==0){//字串没有该字符                left++;                continue;                }                    if(dp[s[left]]>tHash[s[left]]){                        dp[s[left]]--;                        left++;                    }else{                        break;                    }                }                if(right-left+1<min){                    min=right-left+1;                    start=left;                }            }        }                if(min==INT_MAX)//未找到            return "";        //string res=s.substr(start,min); //三种赋值res的方法都可以        //string res(s,start,min);        string res;        for(int i=start;i<start+min;i++)            res+=s[i];        return res;    }};