学步园

转自：

http://www.cnblogs.com/remlostime/archive/2012/11/16/2774077.html

这篇文章的代码是我看到写得最简洁的。

class Solution {
 private:
     int count1[256];
     int count2[256];
 public:
     string minWindow(string S, string T) {
         // Start typing your C/C++ solution below
         // DO NOT write int main() function
         if (T.size() == 0 || S.size() == 0)
             return "";
             
         memset(count1, 0, sizeof(count1));
         memset(count2, 0, sizeof(count2));
         
         for(int i = 0; i < T.size(); i++)
         {
             count1[T[i]]++;
             count2[T[i]]++;
         }
         
         int count = T.size();
         
         int start = 0;
         int minSize = INT_MAX;
         int minStart;
         for(int end = 0; end < S.size(); end++)
         {
             if (count2[S[end]] > 0)
             {
                 count1[S[end]]--;
                 if (count1[S[end]] >= 0)
                     count--;
             }
             
             if (count == 0)
             {
                 while(true)
                 {
                     if (count2[S[start]] > 0)
                     {
                         if (count1[S[start]] < 0)
                             count1[S[start]]++;
                         else
                             break;
                     }
                     start++;
                 }
             
                 if (minSize > end - start + 1)
                 {
                     minSize = end - start + 1;
                     minStart = start;
                 }
             }
         }
         
         if (minSize == INT_MAX)
             return "";
         
         string ret(S, minStart, minSize);
         
         return ret;        
     }
 };