[LeetCode] Group Anagrams

最新推荐文章于 2025-08-20 17:28:47 发布

转载最新推荐文章于 2025-08-20 17:28:47 发布 · 56 阅读

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原文链接：https://yq.aliyun.com/articles/331425

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#开发工具

本文介绍了一种使用 C++ 实现的高效算法来解决将字符串数组中的同字母异序词进行分组的问题。该算法利用了哈希表与多集合的数据结构，并通过排序或计数排序来提高字符串比较效率。

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The function signature has been updated to return a more intuitive vector<vector<string>>which treats a single string as a group of anagrams consisting of only itself.

The idea is to use an unordered_map to store those strings that are anagrams. We use the sorted string as the key and the string itself as the value. The strings are stored in a multisetsince there may be duplicates. Moreover, multiset will sort them by default as we desire.

The code is as follows.

 1 class Solution {
 2 public:
 3     vector<vector<string>> groupAnagrams(vector<string>& strs) {
 4         unordered_map<string, multiset<string>> mp;
 5         for (string s : strs) {
 6             string t = s; 
 7             sort(t.begin(), t.end());
 8             mp[t].insert(s);
 9         }
10         vector<vector<string>> anagrams;
11         for (auto m : mp) { 
12             vector<string> anagram(m.second.begin(), m.second.end());
13             anagrams.push_back(anagram);
14         }
15         return anagrams;
16     }
17 };

Update: general sort takes O(nlogn) time. In this problem, since the string only contains lower-case alphabets, we can write a sorting function using counting sort (O(n) time) to speed up the sorting process. I write a string sorting functionstrSort below and using it to sort the string achieves the overall running time 72ms for this problem while the above code takes 76ms.

 1 class Solution {
 2 public:
 3     vector<vector<string>> groupAnagrams(vector<string>& strs) {
 4         unordered_map<string, multiset<string>> mp;
 5         for (string s : strs) {
 6             string t = strSort(s);
 7             mp[t].insert(s);
 8         }
 9         vector<vector<string>> anagrams;
10         for (auto m : mp) { 
11             vector<string> anagram(m.second.begin(), m.second.end());
12             anagrams.push_back(anagram);
13         }
14         return anagrams;
15     }
16 private:
17     string strSort(string& s) {
18         int count[26] = {0}, n = s.length();
19         for (int i = 0; i < n; i++)
20             count[s[i] - 'a']++;
21         int p = 0;
22         string t(n, 'a');
23         for (int j = 0; j < 26; j++)
24             for (int i = 0; i < count[j]; i++)
25                 t[p++] += j;
26         return t;
27     } 
28 };