leetcode--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.

code 1

public class Solution {
    public String minWindow(String S, String T) {
       if(T.length() == 0 || S.length() < T.length())
			return "";
		Map<Character, Integer> mult = new HashMap<Character, Integer>();
		Map<Character, Integer> alreadyRead = new HashMap<Character, Integer>();
		for(int i = 0; i < T.length(); ++i){
			if(mult.containsKey(T.charAt(i)))
				mult.put(T.charAt(i), mult.get(T.charAt(i)) + 1);
			else {
				mult.put(T.charAt(i), 1);
				alreadyRead.put(T.charAt(i), 0);
			}
		}
		
		String minWindow = "";
		int cnt = 0, start = 0, end = 0, currentLength = S.length() + 1;
		for(; end < S.length(); ++end){
			if(!mult.containsKey(S.charAt(end)))
				continue;
			else {
				int num = alreadyRead.get(S.charAt(end));
				alreadyRead.put(S.charAt(end), num + 1);
				if(num + 1 <= mult.get(S.charAt(end)))
					++cnt;
			}
			
			if(cnt == T.length()){
				while(!mult.containsKey(S.charAt(start)) || alreadyRead.get(S.charAt(start)) > mult.get(S.charAt(start))){
				    if(mult.containsKey(S.charAt(start))){
					    int num = alreadyRead.get(S.charAt(start));
					    alreadyRead.put(S.charAt(start), num - 1);
				    }
					++start;
				}
				
				if(end - start + 1 < currentLength) {
					minWindow = S.substring(start, end + 1);
					currentLength = end - start + 1;
				}
			}
		}
		return minWindow;
    }
}

　　

the following two codes follow the same algorithm with different data structures. Both are accepted

public class Solution {
    public String minWindow(String S, String T) {
       String result = "";
		int slen = S.length(), tlen = T.length();
		if(slen > 0 && tlen > 0 && slen >= tlen){
			HashMap<Character, Integer> mult = new HashMap<Character, Integer>();
			for(int i = 0; i < tlen; ++i){
				int mu = 1;
				if(mult.containsKey(T.charAt(i)))
					mu = mult.get(T.charAt(i)) + 1;
				mult.put(T.charAt(i), mu);
			}
			//find the substring of S which contains all characters in T
			int size = mult.size();
			int start = 0, end = 0;
			for(int i = 0; i < slen; ++i){
				if(mult.containsKey(S.charAt(i))){
					int mu = mult.get(S.charAt(i)) - 1;
					if(mu == 0)
						--size;
					mult.put(S.charAt(i), mu);
				}
				if(size == 0)
					break;
				++end;
			}
			
			if(size == 0){				
				start = shorten(start, end, mult, S);
				result = S.substring(start, end + 1);
				int len = end - start + 1, newend = end;
				while(newend + 1 < slen){
					++newend;
					if(mult.containsKey(S.charAt(newend))){
						int mu = mult.get(S.charAt(newend)) - 1;
						mult.put(S.charAt(newend), mu);
						start = shorten(start, newend, mult, S);
						if(newend - start + 1 < len){
							result = S.substring(start, newend + 1);
							len = newend - start + 1;
							end = newend;
						}
					}
				}
			}			
		}
		return result;       
    }
	
	public int shorten(int start, int end, HashMap<Character, Integer> mult, String S){
		while(start < end){
			if(!mult.containsKey(S.charAt(start)))
				++start;
			else{
				if(mult.get(S.charAt(start)) < 0){
					int j = mult.get(S.charAt(start)) + 1;
					mult.put(S.charAt(start), j);
					++start;							
				}	
				else
					break;
			}	
		}
		return start;
    }
}

　　　

Another code

public class Solution {
    public String minWindow(String S, String T) {
        String result = "";
		int slen = S.length(), tlen = T.length();
		if(slen > 0 && tlen > 0 && slen >= tlen){
			int[] target = new int[256];
			int distChar = 0;
			for(int i = 0; i < tlen; ++i){
				++target[T.charAt(i)];
				distChar = (target[T.charAt(i)] == 1)? distChar + 1 : distChar;
			}
			
			int start = 0, end = 0;
			int[] source = new int[256];			
			for(int i = 0; i < slen; ++i){
				if(target[S.charAt(i)] != 0){
					++source[S.charAt(i)];
					distChar = (source[S.charAt(i)] == target[S.charAt(i)]) ? distChar - 1: distChar; 
				}
				if(distChar == 0)
					break;
				++end;
			}
			if(distChar == 0){
				start = shorten(target, source, start, end, S);
				result = S.substring(start, end + 1);
				int len = end - start + 1, newend = end;
				while(newend + 1 < slen){
					++newend;
					if(target[S.charAt(newend)] != 0){
						++source[S.charAt(newend)];
						start = shorten(target, source, start, newend, S);
						if(newend - start + 1 < len){
							len = newend - start + 1;
							end = newend;
							result = S.substring(start, end + 1);
						}
					}
				}
			}		
		}
		return result;
	}
	
	public static int shorten(int[] target, int[] source, int start, int end, String S){
		while(start < end){
			if(target[S.charAt(start)] == 0)
				++start;
			else{ 
				if(source[S.charAt(start)] > target[S.charAt(start)]){
					--source[S.charAt(start)];
					++start;
				}
				else
					break;
			}			
		}
		return start;
    }
}