Substring with Concatenation of All Words

最新推荐文章于 2024-01-06 14:27:05 发布

原创最新推荐文章于 2024-01-06 14:27:05 发布 · 550 阅读

0 ·

CC 4.0 BY-SA版权

Other Algorithms 同时被 2 个专栏收录

66 篇文章

订阅专栏

Medium

29 篇文章

订阅专栏

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特定单词组成的子串的所有起始索引位置，适用于所有相同长度的单词列表。通过使用哈希映射来存储单词及其出现次数，算法能快速地检查子串是否符合要求。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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).

public class Solution {
    public ArrayList<Integer> findSubstring(String S, String[] L) {
		// Start typing your Java solution below
		// DO NOT write main() function
		ArrayList<Integer> res = new ArrayList<Integer>();
		int words_num = L.length;
		if(words_num <= 0)
			return res;
		HashMap<String, Integer> words = new HashMap<String, Integer>();
		HashMap<String, Integer> curString = new HashMap<String, Integer>();
		for(int i = 0; i < words_num; i++){
			if(words.get(L[i]) != null)
				words.put(L[i], words.get(L[i]) + 1);
			else
				words.put(L[i], 1);
		}
		int len = L[0].length();
		for(int i = 0; i <= S.length() - words_num * len; i++){// i 为S截取部分的起始坐标，末尾不及words_num * len长度的不用比较
			curString.clear();
			for(int j = 0; j < words_num; j++){
				String cur = S.substring(i + j * len, i + j * len + len);// 截取len长度的词
				if(!words.containsKey(cur))// 比较
					break;
				if(curString.get(cur) != null)// 统计个数
					curString.put(cur, curString.get(cur) + 1);
				else
					curString.put(cur, 1);
				if(curString.get(cur) > words.get(cur))// 超出个数
					break;
				if(j == words_num - 1)// 全部词汇用完
					res.add(i);
			}
		}
		return res;
	}
}

每次截取S的一部分逐个比较L中的词