[LeetCode] Minimum Window Substring

Problem:

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.

Solution:O(m*n)

public class Solution {
    public String minWindow(String S, String T) {  
    if(S == null || T == null || S.length()==0 || T.length()==0)  
        return "";  
    HashMap<Character, Integer> map = new HashMap<>();  
    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 ret = "";  
    int minLen = S.length()+1;  
    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(minLen>i-pre+1)  {  
                            ret = S.substring(pre,i+1);  
                            minLen = i-pre+1;  
                        }  
                        count--;  
                    }  
                }  
                pre++;  
            }  
        }  
    }  
    return ret;  
  }  
}