Minimum Window Substring leetcode java

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 emtpy string

""

.

If there are multiple such windows, you are guaranteed that there will always be only one unique minimum window in S.

题解：

这道题也是用滑动窗口的思想，思想跟 Substring with Concatenation of All Words是一样的，同样是利用HashMap来存Dict，然后来遍历整个母串。因为这里是要求最短的包含子串的字符串，所以中间是可以允许有非子串字符的，当遇见非子串字符而count又没到子串长度时，可以继续走。

当count达到子串长度，说明之前遍历的这些有符合条件的串，用一个pre指针帮忙找，pre指针帮忙找第一个在HashMap中存过的，并且找到后给计数加1后的总计数是大于0的，判断是否为全局最小长度，如果是，更新返回字符串res，更新最小长度，如果不是，继续找。

这道题的代码也参考了code ganker的。

代码如下：

1 public String minWindow(String S, String T) {
2 String res = "";
3 if(S == null || T == null || S.length()==0 || T.length()==0)
4 return res;
5
6 HashMap<Character, Integer> dict = new HashMap<Character, Integer>();
7 for(int i =0;i < T.length(); i++){
8 if(!dict.containsKey(T.charAt(i)))
9 dict.put(T.charAt(i), 1);
else
dict.put(T.charAt(i), dict.get(T.charAt(i))+1);
}

int count = 0;
int pre = 0;
int minLen = S.length()+1;
for(int i=0;i<S.length();i++){
if(dict.containsKey(S.charAt(i))){
dict.put(S.charAt(i),dict.get(S.charAt(i))-1);
if(dict.get(S.charAt(i)) >= 0)
count++;

while(count == T.length()){
if(dict.containsKey(S.charAt(pre))){
dict.put(S.charAt(pre),dict.get(S.charAt(pre))+1);

if(dict.get(S.charAt(pre))>0){
if(minLen>i-pre+1){
res = S.substring(pre,i+1);
minLen = i-pre+1;
}
count--;
}
}
pre++;
}
}//end for if(dict.containsKey(S.charAt(i)))
}
return res;
}Reference：
http://blog.csdn.net/linhuanmars/article/details/20343903