Leetcode之Substring with Concatenation of All Words

最新推荐文章于 2025-09-05 23:16:08 发布

原创最新推荐文章于 2025-09-05 23:16:08 发布 · 156 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode

leetcode和机试题专栏收录该内容

509 篇文章

订阅专栏

本文介绍了一种高效的字符串匹配算法，用于在给定字符串中查找由多个相同长度单词组成的子串的所有起始索引。通过使用map数据结构，算法能够显著降低时间复杂度，实现快速查找。

题目：

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 = "wordgoodgoodgoodbestword",
  words = ["word","good","best","word"]
Output: []

代码：

class Solution {
public:

 vector<int> findSubstring(string s, vector<string>& words) {
      vector<int> res;
        if (s.empty() || words.empty()) return res;
        int n = words.size(), m = words[0].size();
        unordered_map<string, int> m1;
        for (auto &a : words) ++m1[a];
        for (int i = 0; i <= (int)s.size() - n * m; ++i) {
            unordered_map<string, int> m2;
            int j = 0; 
            for (j = 0; j < n; ++j) {
                string t = s.substr(i + j * m, m);
                if (m1.find(t) == m1.end()) break;
                ++m2[t];
                if (m2[t] > m1[t]) break;
            }
            if (j == n) res.push_back(i);
        }
        return res;
    }
};

注意：

学会想到利用map来减少时间复杂度