[leetcode] 76. Minimum Window Substring

Description

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).

Example:

Input: S = "ADOBECODEBANC", T = "ABC"
Output: "BANC"

Note:

• If there is no such window in S that covers all characters in T, return the empty string “”.
• If there is such window, you are guaranteed that there will always be only one unique minimum window in S.

分析

题目的意思是：在S中找出最小的窗口，使得该窗口包含T中的所有字符，复杂度为O(n).

• 开始left=0，right向后移动，Tmap是T的字符计数键值对.
• 我们用一个count来计算截取的子串是否合法，当count==T.size，说明截取的子串已经包含所给定的子串，然后我们移动left找最小的子串

代码

class Solution {
public:
    string minWindow(string s, string t) {
        string result;
        if(!s.size()||!t.size()){
            return result;
        }
        int left=0;
        map<char,int> Tmap;
        for(int i=0;i<t.size();i++){
            Tmap[t[i]]++;
        }
        int min_len=s.size()+1;
        int count=0;
        for(int right=0;right<s.size();right++){
            if(Tmap.find(s[right])!=Tmap.end()){
                if(Tmap[s[right]]>0){
                    count++;
                }
                Tmap[s[right]]--;
                while(count==t.size()){
                    if(Tmap.find(s[left])!=Tmap.end()){
                        if(min_len>right-left+1){
                            min_len=right-left+1;
                            result=s.substr(left,right-left+1);
                        }
                        Tmap[s[left]]++;
                        if(Tmap[s[left]]>0){
                            count--;
                        }   
                    }
                    left++;
                }
            }
        }
        return result;
    }
};

参考文献

​​[编程题]minimum-window-substring​​