[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.

思考：虽然也想到了时间换空间的方法，但是时间复杂度还是不能压到O(n)。我的思路是二维数组记录T中每个字符的index，然后转化为一个求最短距离。

参考：http://discuss.leetcode.com/questions/97/minimum-window-substring

此方法的精妙之处在于count！

class Solution {
public:
    string minWindow(string S, string T) {
        int p1[256]={0}; //need to find
        int p2[256]={0}; //has found
        int count=0; 
        int len1=S.size();
        int len2=T.size();
        for(int i=0;i<len2;i++) p1[T[i]]++;
        int start=0,end=0;
        int len=INT_MAX; //window大小
        int pos=-1;
        for(;end<len1;end++)
        {
            if(p1[S[end]]==0) continue; //S[end]不在T中，跳过
            p2[S[end]]++;
            if(p2[S[end]]<=p1[S[end]]) count++;
            if(count==len2) //找到一个window
            {
                while(p1[S[start]]==0||p2[S[start]]>p1[S[start]]) 
                {
                    if(p2[S[start]]>p1[S[start]]) p2[S[start]]--;
                    start++;
                }
                if(len>end-start+1) 
                {
                    len=end-start+1;
                    pos=start;
                }
            }
        }
        return (pos==-1)?"":S.substr(pos,len);
    }
};