76. 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.

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给出了；两个字符串s和t，求s中包含t所有字符的最短的子字符串。要注意的是t中的字符可能是重复的。这里使用两个索引left和right指定符合条件的子字符串的开头和结尾。首先对right自增，直到t中的所有字符都有了。因为有重复的字符，所以用map来记录好重复次数，当所有次数都满足时停止right的增长。现在s[left, right]已经是符合要求的子字符串。为了求最短的，将left增长，直到刚好有一个字符的数量不满足次数要求，现在的s[left, right]就是当前最短的答案。然后又将right增长，求另外的有可能的答案。。。最后返回最短的答案即可。

代码：

class Solution{public:	string minWindow(string s, string t) 	{		int cnt[256] = {0}, map[256] = {0};		int types = 0, left = 0, right = 0, n = 0, len = INT_MAX;		string res;		for(auto c:t) 		{			if(map[c]++ == 0) n++;		}		while(right < s.size())		{			while(right < s.size() && types < n)			{				if(map[s[right]] > 0 && ++cnt[s[right]] == map[s[right]])				{					++types;				}				++right;			}			if(types < n) break;			while(left < right)			{				if(map[s[left]] > 0 && --cnt[s[left]] < map[s[left]])				{					--types;					break;				}				++left;			}			if(right - left < len)			{				len = right - left;				res = s.substr(left, right-left);			}			++left;		}		return res;	}};
在别人的答案中看到一个号称是最短的答案，挺不错的：
string minWindow(string s, string t) {        vector<int> map(128,0);        for(auto c: t) map[c]++;        int counter=t.size(), begin=0, end=0, d=INT_MAX, head=0;        while(end<s.size()){            if(map[s[end++]]-->0) counter--; //in t            while(counter==0){ //valid                if(end-begin<d)  d=end-(head=begin);                if(map[s[begin++]]++==0) counter++;  //make it invalid            }          }        return d==INT_MAX? "":s.substr(head, d);    }