Substring with Concatenation of All Words

问题描述

You are given a string, s, and a list of words, words, that are all of the same length. Find all starting indices of substring(s) in s that is a concatenation of each word in wordsexactly once and without any intervening characters.

For example, given:
s: "barfoothefoobarman"
words: ["foo", "bar"]

You should return the indices: [0,9].
(order does not matter).

解决思路

最直观的办法：

1. 用一个boolean[] flag，大小和输入的字符串相同，flag[i]代表s[i...i+lenOfToken]是否在words中；

2. 计算出拼接字符串的长度，然后依次检查是否完全覆盖words（用一个HashMap记录）。

时间复杂度为O(n)，空间复杂度为O(n).

注意：words中的word有可能有重复。

程序

public class Solution {
    public List<Integer> findSubstring(String s, String[] words) {
		List<Integer> res = new ArrayList<>();
		if (s == null || s.length() == 0 || words == null || words.length == 0) {
			return res;
		}
        
                // <word, count>
		HashMap<String, Integer> map = new HashMap<String, Integer>(); 
		for (String w : words) {
			if (map.containsKey(w)) {
				map.put(w, map.get(w) + 1);
			} else {
				map.put(w, 1);
			}
		}

		int lenOfToken = words[0].length();
		int numOfToken = words.length;
		int len = s.length();
		boolean[] flag = new boolean[len]; // speed up

		int i = 0;
		while (i + lenOfToken <= len) {
			String sub = s.substring(i, i + lenOfToken);
			if (map.containsKey(sub)) {
				flag[i] = true;
			}
			++i;
		}

		int totalLen = lenOfToken * numOfToken;
		for (i = 0; i + totalLen <= len; i++) {
			if (!flag[i]) {
				continue;
			}
			int k = numOfToken;
			int j = i;
			HashMap<String, Integer> map_tmp = new HashMap<String, Integer>(map);
			while (k > 0) {
				String word_tmp = s.substring(j, j + lenOfToken);
				if (!flag[j] || !map_tmp.containsKey(word_tmp)
						|| map_tmp.get(word_tmp) == 0) {
					break;
				}
				map_tmp.put(word_tmp, map_tmp.get(word_tmp) - 1);
				j += lenOfToken;
				--k;
			}
			if (k == 0) {
				res.add(i);
			}
		}

		return res;
	}
}