76. Minimum Window Substring (String, Map)

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：问题可以转变为，T中各个字符的数量<= window中的这些字符的数量，所以用map纪录字符与字符数量的关系

class Solution {
public:
    bool ifContain(map<char,int>& source, map<char,int>& target) //check if S contains T
    {
        for(map<char,int>::iterator it = target.begin(); it != target.end(); it++) //map的遍历
        {
            if(source[it->first] < it->second) return false;
        }
        return true;
    }
 
    string minWindow(string S, string T) {
        int minLength = INT_MAX;
        int start = , end = , minStart = , minEnd = ;
        map<char,int> source;
        map<char,int> target;
        source[S[start]]++; //map的插入[法I]source[key]=value; [法II]source.insert(make_pair(key,value));
        for(int i = ; i< T.length(); i++)
        {
            target[T[i]]++;
        }
        while()
        {
            if(ifContain(source, target)){
                if(end-start+ < minLength)
                {
                    minStart = start;
                    minEnd = end;
                    minLength = end-start+;
                    if(minLength == T.size()) return S.substr(minStart,minLength);
                }
                source[S[start]]--; //寻找更小的窗口
                start++;
            }
            else //不包含，则扩大窗口
            {
                end++;
                if(end==S.size()) break;
                source[S[end]]++;
            }
        }
        if(minLength>S.size()) return "";
        else return S.substr(minStart,minLength);
    }
};