LeetCode | Minimum Window Substring

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.

采用双指针遍历，尾指针遍历至包含T的所有字符，首指针尽可能向后压缩，然后在所有窗口中记录最小窗口位置。

参考http://leetcode.com/2010/11/finding-minimum-window-in-s-which.html

代码

public class MinimumWindowSubstring {
	public String minWindow(String S, String T) {
		if (T == null || T.length() == 0 || S == null || S.length() == 0) {
			return "";
		}

		int M = S.length();
		int N = T.length();
		char[] s = S.toCharArray();
		char[] t = T.toCharArray();

		// use as map
		int[] needToFind = new int[256];
		int[] hasFound = new int[256];
		for (int i = 0; i < N; ++i) {
			++needToFind[t[i]];
		}
		int count = 0;
		int minWinSize = Integer.MAX_VALUE;
		int left = 0;
		int right = 0;
		for (int begin = 0, end = 0; end < M; ++end) {
			if (needToFind[s[end]] == 0) {
				continue;
			}
			++hasFound[s[end]];
			if (hasFound[s[end]] <= needToFind[s[end]]) {
				++count;
			}
			if (count == N) {
				while (needToFind[s[begin]] == 0
						|| (hasFound[s[begin]] > needToFind[s[begin]])) {
					if ((hasFound[s[begin]] > needToFind[s[begin]])) {
						--hasFound[s[begin]];
					}
					++begin;
				}
				int winSize = end - begin + 1;
				if (winSize < minWinSize) {
					minWinSize = winSize;
					left = begin;
					right = end;
				}
			}
		}
		if (count == N) {
			return S.substring(left, right + 1);
		}
		return "";
	}
}