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.
题目的意思是说，要在S中找一个最短的子串，该子串包含了T中的所有字符，如上面所示<span style="font-size:18px;">class Solution {
public:
string minWindow(string S, string T) {
int totalnumber[256]={0};
int aredyfind[256]={0};
int count=0;//用来表示当前已经找到的总字符，如果都找到，count为T.length()
int i,start;
int winstart=-1;
int winend=S.length();
for(i=0;i<T.size();i++)
{
totalnumber[T[i]]++;
}
for(i=0,start=0;i<S.size();i++)
{
if(totalnumber[S[i]]!=0) //这里表示字符S[i]属不属于T，如果不属于T那么直接跳过
{
aredyfind[S[i]]++;
if(aredyfind[S[i]]<=totalnumber[S[i]])
count++;
if(count==T.length()) //表示已经找到了一个合法的窗口
{
while(totalnumber[S[start]]==0||aredyfind[S[start]]>totalnumber[S[start]]) //这里判断start处的字符属不属于T或者属于T但是重复出现了，两种情况都是要缩小窗口的
{
if(aredyfind[S[start]]!=0)
aredyfind[S[start]]--;
start++;
}
if(winend-winstart>i-start)
{
winend=i;
winstart=start;
}
aredyfind[S[start]]--; //上面表示已经找到了一个start开始 i结束的当前最优，然后start++，此时必然有字母被删除了，所以回到开始i出循环，i往后遍历来寻找哪个被删除的字母
start++;
count--;
}
}
}
return winstart==-1?"":S.substr(winstart,winend-winstart+1);
}
};</span>Minimum Window Substring leetcode
原文地址：http://blog.csdn.net/yujin753/article/details/41809705
