标签：leetcode   算法   

问题描述：

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 emtpy string "".

If there are multiple such windows, you are guaranteed that there will always be only one unique minimum window in S.

代码：

 string minWindow(string S, string T) {  //C++
        int size_s = S.size();
        int size_t = T.size();
        if(size_t == 0||size_s==0||size_s<size_t)
            return "";
        
        string result;
        map<char,int> dic;
        map<char,int> window;
        int validNum = 0,min = INT_MAX;
        
        for(int i=0; i<size_t; i++)
            dic[T[i]]++;
        for(int fast=0,slow=0; fast<size_s;fast++){
            char ch = S[fast];
            if(dic.count(ch)!=0){
                window[ch]++;
                if(window[ch] <= dic[ch])
                    validNum++;
            }
            if(validNum == size_t){
                while(dic.count(S[slow])==0||dic[S[slow]]<window[S[slow]]){
                    if(window.count(S[slow])!=0)
                        window[S[slow]]--;
                    slow++;
                }
                if(fast-slow+1 < min){
                    min = fast-slow+1;
                    result = S.substr(slow,min);
               }
            }
            
        }
        return result;
    }

[leetcode] 76 Minimum Window Substring

标签：leetcode   算法   

原文地址：http://blog.csdn.net/chenlei0630/article/details/43232697