leetcode[76]Minimum Window Substring

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.

class Solution {
public:
/*
string minWindow(string S, string T) 
{
    string res="";
    if(S.empty()||T.empty()||S.size()<T.size())return res;
    int countNeed[256]={0};
    int countHas[256]={0};
    int countT=T.size();
    for (int i=0;i<countT;i++)
    {
        countNeed[T[i]]++;
        countHas[T[i]]++;
    }
    int start=0;
    int minSize=INT_MAX;
    for (int end=0;end<S.size();end++)
    {
        if (countHas[S[end]]>0)
        {
            countNeed[S[end]]--;
            if(countNeed[S[end]]>=0)countT--;
        }
        if (countT==0)
        {
            while(start<S.size())
            {
                if (countHas[S[start]]>0)
                {
                    if (countNeed[S[start]]<0)
                    {
                        countNeed[S[start]]++;
                        start++;
                    }
                    else  break;
                }
                else  start++;
            }
            if (end-start+1<=minSize)
            {
                minSize=end-start+1;
                res=S.substr(start,minSize);
            }
        }
    }
    return res;
}
*/

string minWindow(string S, string T) 
{
    string res="";
    if(S.empty()||T.empty()||S.size()<T.size())return res;
    int countT=T.size();
    map<char,int> MapNeed;
    for (int i=0;i<countT;i++)
    {
        MapNeed[T[i]]=0;
    }
    for (int i=0;i<countT;i++)
    {
        MapNeed[T[i]]++;
    }
    int start=0;
    int minSize=INT_MAX;
    for (int end=0;end<S.size();end++)
    {
        if (MapNeed.find(S[end])!=MapNeed.end())
        {
            MapNeed[S[end]]--;
            if (MapNeed[S[end]]>=0)countT--;
        }
        if (countT==0)
        {
            while (start<S.size())
            {
                if (MapNeed.find(S[start])!=MapNeed.end())
                {
                    if (MapNeed[S[start]]<0)
                    {
                        MapNeed[S[start]]++;
                        start++;
                    } 
                    else
                    {
                        break;
                    }
                }
                else  
                {
                    start++;
                }
            }
            if (end-start+1<=minSize)
            {
                minSize=end-start+1;
                res=S.substr(start, minSize);
            }
        }
    }
    return res;
}

};