76. Minimum Window Substring

标签（空格分隔）： leetcode hard

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题目

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.

思路

本题是典型的子字符串求解问题。对于大多数子字符串问题，我们得到一个字符串，需要找到满足某些限制条件的子字符串。一般的方法是使用一个包含两个指针的hashmap。模板如下所示。

int findSubstring(string s){        vector<int> map(128,0);        int counter; // check whether the substring is valid        int begin=0, end=0; //two pointers, one point to tail and one  head        int d; //the length of substring        for() { /* initialize the hash map here */ }        while(end<s.size()){            if(map[s[end++]]-- ?){  /* modify counter here */ }            while(/* counter condition */){                  /* update d here if finding minimum*/                //increase begin to make it invalid/valid again                if(map[s[begin++]]++ ?){ /*modify counter here*/ }            }              /* update d here if finding maximum*/        }        return d;  }

本题的解决方案就呼之欲出了

Note：需要提到的一点是，当要求找到最大子字符串时，我们应该在内部while循环之后更新最大值，以保证子字符串是有效的。另一方面，当要求找到最小子字符串时，我们应该在内部while循环中更新最小值。

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代码

class Solution {public:    string minWindow(string s, string t) {        //定义变量count,头尾指针begin及end        int count=t.size(),begin=0,end=0,head=0,d=INT_MAX;        //利用ASCII码来代表字符并进行存储        vector<int> map(128,0);        for(auto item:t)            map[item]++;        //移动头尾指针开始寻找最小子串        while(end<s.size())        {            if(map[s[end++]]-->0)                count--;            while(count==0)            {                //比较是否为最短子串                if((end-begin)<d) d =end-(head=begin);                if(map[s[begin++]]++==0)                    count++;            }        }        return d==INT_MAX?"":s.substr(head,d);            }};