Substring with Concatenation of All Words

最新推荐文章于 2024-01-06 14:27:05 发布

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本文介绍了一种高效的字符串匹配算法，该算法能在给定的字符串中找到由多个指定单词组成的子串的所有起始位置。通过使用哈希映射来记录目标单词出现的次数，此算法能够有效地检查每个可能的子串是否符合要求。

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原题：

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

For example, given:
S: "barfoothefoobarman"
L: ["foo", "bar"]

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

解答：

public List<Integer> findSubstring(String S, String[] L) {
		ArrayList<Integer> result = new ArrayList<Integer>();
		if(S==null || S.isEmpty() || L==null || L.length <1) {
			return result;
		}
		HashMap<String, Integer> toFind = new HashMap<String, Integer>();
		HashMap<String, Integer> find = new HashMap<String, Integer>();
		int arr_len = L.length;
		int str_len = L[0].length();
		for(int i=0; i<arr_len; i++) {
			if(!toFind.containsKey(L[i])) {
				toFind.put(L[i], 1);
			} else {
				toFind.put(L[i], toFind.get(L[i])+1);
			}
		}
		for(int i=0; i<=S.length()-arr_len*str_len; i++) {
			find.clear();
			int j;
			for(j=0; j<arr_len; j++) {
				int k = i+j*str_len;
				String subStr = S.substring(k, k+str_len);
				if(!toFind.containsKey(subStr)) break;
				if(!find.containsKey(subStr)) {
					find.put(subStr, 1);
				} else {
					find.put(subStr, find.get(subStr)+1);
				}
				if(find.get(subStr) > toFind.get(subStr)) break;
			}
			if(j == arr_len) result.add(i);
		}
		return result;
    }