Minimum Window Substring

来源：Leetcode

原帖：http://oj.leetcode.com/problems/minimum-window-substring/

题目：
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.
Solution: Use two pointers: start and end.
               First, move 'end'. After finding all the needed characters, move 'start'.
               Use array as hashtable.

代码：

 class Solution {  
 public:  
   string minWindow(string S, string T) {  
     int N = S.size(), M = T.size();  
     if (N < M) return "";  
     int need[256] = {0}, find[256] = {0};  
     for (int i = 0; i < M; ++i) need[T[i]]++;        
     int count = 0;   
     int resStart = -1, resEnd = N;  
     for (int start = 0, end = 0; end < N; ++end) {  
       if (need[S[end]] == 0) continue;    
       if (++find[S[end]] <= need[S[end]]) count++;  
       if (count != M) continue;       
       // move 'start'  
       for (; start < end; ++start) {  
         if (need[S[start]] == 0) continue;  
         if (find[S[start]] == need[S[start]]) break;  
         find[S[start]]--;  
       }        
       // update min window  
       if (end - start < resEnd - resStart) {  
         resStart = start;  
         resEnd = end;  
       }  
     }  
     return (resStart == -1) ? "" : S.substr(resStart, resEnd - resStart + 1);  
   }  
 };