LeetCode--minimum-window-substring

题目描述

Given a string S and a string T, find the minimum window
7646
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：最小滑动窗口问题，利用左右指针，首先找到最大的包含T所有字符的窗口，然后将左指针向右移动，直到找到在T中存在的字符，并且满足题目要求。然后再将右指针向右移动，将找到的最小滑动窗口与之前的窗口比较，从而找到整体最小的窗口。
class Solution {
public:
    //维持一个窗口滑动，左边是left，右边是right，然后判断是否包含T
    string minWindow(string S, string T)
    {
        string result;
        if(!S.size() || !T.size())
        {
            return result;
        }
        map<char, int>Tmap;   //存储T字符串里字符，便于与S匹配
        int left = 0;       //左边窗口,起始为0
        int count = 0;      //计数器，对窗口内T串字符数量进行计数,起始为0
        //装载T串
        int minlen = S.size() + 1;      //最小窗口，便于比较最后取最小的,初始化取最大
        for(int i = 0; i < T.size(); ++i)
        {
            Tmap[T[i]]++;
        }
        //移动右边窗口
        for(int right = 0; right < S.size(); ++right)
        {
            if(Tmap.find(S[right]) != Tmap.end())   //当窗口内部有T中的字符
            {
                if(Tmap[S[right]] > 0)
                {
                    count++;            //计数器+1
                }
                Tmap[S[right]]--;   //去掉包含的，剩下都是没包含的
                while(count == T.size())//当窗口内有T的所有字符，就要开始移动左窗口啦
                {
                    if(Tmap.find(S[left]) != Tmap.end())
                    {
                        //好了，这就是一个包含字符串的窗口范围：left ~ right,
                        //判断是否比当前窗口还小，是就取串
                        if(minlen > right - left + 1)
                        {
                            //更新窗口大小
                            minlen = right -left + 1;
                            result = S.substr(left, right - left + 1);
                        }
                        //舍弃窗口左边字符串，继续移动窗口
                        Tmap[S[left]]++;
                        if(Tmap[S[left]] > 0)  //如果左边连续相同，则count不递减，窗口需要继续向右移动
                        {
                            count--;
                        }
                    }
                    left++;             //移动左窗口
                }
            }
        }
        return result;
    }
};