LeetCode，

LeetCode，

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 emtpy string "".

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

分析：

一前一后两个指针，后指针向后寻找到包含T的区间，移动前指针到最优位置，判断一次，如此循环。

class Solution {
public:
    string minWindow(string S, string T) {
        int Need[256], Now[256];
        for(int i = 0;i < 256; i++){
            Need[i] = Now[i] = 0;
        }
        int LS = S.length();
        int LT = T.length();
        for(auto ch:T)
            Need[ch] ++;
        int i = 0, j = 0, foundNum = 0;
        int minLength = LS + 1, Begin = 0, End = 0;
        while(j < LS){
            if(!Need[S[j]]){
                j++;
                continue;
            }
            Now[S[j]]++;
            if(Now[S[j]] <= Need[S[j]])
                ++foundNum;
            if(foundNum == LT){
                while(Need[S[i]] == 0 || Now[S[i]] > Need[S[i]]){
                    if(Now[S[i]] > Need[S[i]])
                        --Now[S[i]];
                    i++;
                }
                int l = j - i + 1;
                if(l < minLength){
                    minLength = l;
                    Begin = i;
                    End = j;
                }

            }
            j++;
        }
        return minLength <= LS ? S.substr(Begin, minLength) :"";
    }
};