[LeetCode] 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.

 这道题开始没什么思路，感觉O(n)这个限制很变态。

后来参考了http://www.cnblogs.com/remlostime/archive/2012/11/16/2774077.html

“双指针思想，尾指针不断往后扫，当扫到有一个窗口包含了所有T的字符，然后再收缩头指针，直到不能再收缩为止。最后记录所有可能的情况中窗口最小的”

class Solution {
public:
    string minWindow(string S, string T) {
        //set<int> Tset;
        int *count1 = new int [270]; //set to 0??
        int *count2 = new int [270];
        
        for (int i=0; i<270; i++)
        {
            count1[i] = count2[i] = 0;
        }
        
        for (int i=0; i<T.size(); i++)
        {
            count1[T[i]]++;
            count2[T[i]]++;
        }
        
        int count = T.size();
        int start = 0, end = 0;
        int minstart = 0;
        int minsize = INT_MAX;
        
        for (end=0; end<S.size(); end++)
        {
            if (count1[S[end]] > 0)
            {
                count2[S[end]]--;
                if (count2[S[end]] >= 0)
                {
                    count--;
                }
            }
            if (count == 0)
            {
                while (true)
                {
                    if (count1[S[start]] > 0)
                    {
                        if (count2[S[start]] < 0) //<0 stand for that character in T in S has more than one.
                        {
                            count2[S[start]]++;
                        }
                        else
                            break;
                    }
                    start++;
                }
                if (end - start + 1 < minsize)
                {
                    minsize = end - start + 1;
                    minstart = start;
                }
            }
        }
        
        
        if (minsize == INT_MAX)
            return "";
        
        string ret(S, minstart, minsize);
        
        return ret;
        
    }
};