30-字串为其他字符串连接所得

最新推荐文章于 2025-08-24 23:47:30 发布

kevin聪

最新推荐文章于 2025-08-24 23:47:30 发布

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分类专栏：算法与数据结构 leetcode 文章标签：算法数据结构 leetcode

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本文介绍了一种用于查找字符串中特定模式的有效算法。该算法能在给定的字符串s中找到所有由words列表中单词按顺序拼接而成的子串，并返回这些子串的起始索引。通过使用两个哈希映射分别记录目标单词出现次数和已匹配单词数量，算法能够高效地处理问题。

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Description

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 words exactly once and without any intervening characters.

Example 1:

Input:
  s = "barfoothefoobarman",
  words = ["foo","bar"]
Output: [0,9]
Explanation: Substrings starting at index 0 and 9 are "barfoor" and "foobar" respectively.
The output order does not matter, returning [9,0] is fine too.

Example 2:

Input:
  s = "wordgoodstudentgoodword",
  words = ["word","student"]
Output: []

问题描述

给定字符串s和字符串列表words(words中所有字符串的长度相等).找出s中的所有起始下标，使得s.substring(i)为words中每个单词的连接(每个单词有且一次)并且不包含任何交织的字符

问题分析

维护两个map

wcount, 存储words中对应单词的个数
seen, 在迭代下标i的过程中维护的map，存储当前见过的在words中的单词对应个数

思想

迭代下标，对每个下标维护seen，令某个在words中的单词为temp，若seen.get(temp) > wcount.get(temp),则终止迭代，否则计数增1，当计数等于words长度时，说明i有效，将i添加入结果中。此过程反复进行

解法

class Solution {
    public List<Integer> findSubstring(String s, String[] words) {
        List<Integer> res = new ArrayList();
        if(s == null || words == null || words.length == 0)  return res;

        int n = s.length(), len = words[0].length(), num = words.length;
        if(n < len * num)   return res;

        Map<String, Integer> wcount = new HashMap();

        for(String word : words)    wcount.put(word, wcount.getOrDefault(word, 0) + 1);

        for(int i = 0;i <= n - num * len;i++){
            Map<String, Integer> seen = new HashMap();
            int j = 0;
            for(;j < num;j++){
                String str = s.substring(i + j * len, i + (j + 1) * len);
                if(wcount.containsKey(str)){
                    seen.put(str, seen.getOrDefault(str, 0) + 1);
                    if(seen.get(str) > wcount.get(str)){
                        break;
                    }
                }else{
                    break;
                }
            }
            if(j == num){
                res.add(i);
            }
        }

        return res;
    }
}