学步园

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.

用hashtable的时候，如果key不存在而取值会报错，多个地方需要检查存在与否，不太合适。不如直接使用int[256]数组来记录。

import java.util.Hashtable;

public class Solution {
    public String minWindow(String S, String T) {
		// Start typing your Java solution below
		// DO NOT write main() function
		Hashtable<Character, Integer> goal = new Hashtable<Character, Integer>();
		Hashtable<Character, Integer> process = new Hashtable<Character, Integer>();
		int lenS = S.length();
		int lenT = T.length();
		int sum = lenT;
		int counter = 0;
		for (int i = 0; i < lenT; i++) {
			char cur = T.charAt(i);
			process.put(cur, 0);
			if (!goal.containsKey(cur))
				goal.put(cur, 1);
			else {
				goal.put(cur, goal.get(cur) + 1);
			}
		}
		int begin = 0, end = 0, minWin = S.length() + 1;
		int min_beg = 0, min_end = 0;
		while (end < lenS) {
			char cur = S.charAt(end);
			char beg_char = S.charAt(begin);
			if (!goal.containsKey(cur) || goal.get(cur) == 0) {
				end++;
				continue;
			}
			process.put(cur, process.get(cur) + 1);
			if (process.get(cur) <= goal.get(cur))
				counter++;
			if (counter == sum) {
				while (!process.containsKey(beg_char) || process.get(beg_char) > goal.get(beg_char)) {
					if(process.containsKey(beg_char)){
						process.put(beg_char, process.get(beg_char) - 1);
					}
					begin++;
					beg_char = S.charAt(begin);
				}
				int len = end - begin + 1;
				if (len < minWin) {
					minWin = len;
					min_beg = begin;
					min_end = end;
				}
			}
			end++;
		}
		return minWin <= lenS ? S.substring(min_beg, min_end + 1) : "";
	}
}