用unordered_map存储t中的字符和存储的次数，l是字符串最左边的字符的位置，r是字符串最右边字符的位置，count是s中从l到r这一区间成功匹配t中字符个数。当count的个数跟t的大小一样大(也就是成功匹配)，就将当前子串的size和min_size比较以更新min_size，会出现一种情况，l位置的字符并不在t中，这样可以继续移动l以求得真正的min_size

class Solution {public:    string minWindow(string s, string t) {        int length1 = s.size();        int length2 = t.size();        if(length1 <= 0 || length2 <= 0)            return "";        unordered_map
     
       container;        for(int i = 0;i < length2;i++)            container[t[i]]++;        int l = 0,count = 0,min_l = 0,min_size = s.size() + 1;        for(int r = 0;r < length1;r++){            if(container.find(s[r]) != container.end()){                if(--container[s[r]] >= 0)                    count++;                while(count == t.size()){                    int size = r - l + 1;                    if(size < min_size){                        min_l = l;                        min_size = size;                    }                    if(container.find(s[l]) != container.end()){                        if(++container[s[l]] > 0)                            count--;                     }                    l++;                }            }        }        if(min_size > s.size())             return "";        return s.substr(min_l, min_size);          }};
     

 https://blog.csdn.net/makuiyu/article/details/44570193