[leetcode] Minimum WIndow

题目：（HashTable Two Point String）

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：

有一个思想叫 滑动窗口：参考http://www.cnblogs.com/springfor/p/3872559.html

public class Solution {
    public String minWindow(String S, String T) {
        if(S.length()==0||T.length()==0||S==null||T==null)
           return "";
           
        HashMap <Character,Integer> map =new  HashMap <Character,Integer>();
        for(int i=0;i<T.length();i++)
            if(map.containsKey(T.charAt(i)))
                map.put(T.charAt(i),map.get(T.charAt(i))+1);
            else
                map.put(T.charAt(i),1);
        
        int count=0;
        int pre=0;
        String res="";
        int min=S.length()+1;//equal = s.length() just set a initial value
        for(int i=0;i<S.length();i++)
        {
            if(map.containsKey(S.charAt(i)))
            {
                 map.put(S.charAt(i),map.get(S.charAt(i))-1);
                 if(map.get(S.charAt(i))>=0)
                   count++;
                 while(count==T.length())
                 {
                       if(map.containsKey(S.charAt(pre)))
                       {
                            map.put(S.charAt(pre),map.get(S.charAt(pre))+1);
                            if(map.get(S.charAt(pre))>0)
                            {
                                 if(min>i-pre+1)
                                 {
                                     min=i-pre+1;
                                     res=S.substring(pre,i+1);
                                 }
                                 count--;
                            }
                       }
                       pre++;
                 }
            }
        }
        return res;
    }
}