README_EN.md

49. Group Anagrams

中文文档

Description

Given an array of strings strs, group the anagrams together. You can return the answer in any order.

An Anagram is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.

 

Example 1:

Input: strs = ["eat","tea","tan","ate","nat","bat"]
Output: [["bat"],["nat","tan"],["ate","eat","tea"]]

Example 2:

Input: strs = [""]
Output: [[""]]

Example 3:

Input: strs = ["a"]
Output: [["a"]]

 

Constraints:

• 1 <= strs.length <= 104
• 0 <= strs[i].length <= 100
• strs[i] consists of lower-case English letters.

Solutions

Python3

class Solution:
    def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
        chars = collections.defaultdict(list)
        for s in strs:
            k = ''.join(sorted(list(s)))
            chars[k].append(s)
        return list(chars.values())

Java

class Solution {
    public List<List<String>> groupAnagrams(String[] strs) {
        Map<String, List<String>> chars = new HashMap<>();
        for (String s : strs) {
            char[] t = s.toCharArray();
            Arrays.sort(t);
            String k = new String(t);
            if (!chars.containsKey(k)) {
                chars.put(k, new ArrayList<>());
            }
            chars.get(k).add(s);
        }
        return new ArrayList<>(chars.values());
    }
}

C++

class Solution {
public:
    vector<vector<string>> groupAnagrams(vector<string> &strs) {
        unordered_map<string, vector<string>> chars;
        for (auto s : strs)
        {
            string k = s;
            sort(k.begin(), k.end());
            chars[k].emplace_back(s);
        }
        vector<vector<string>> res;
        for (auto it = chars.begin(); it != chars.end(); ++it)
        {
            res.emplace_back(it->second);
        }
        return res;
    }
};

Go

func groupAnagrams(strs []string) [][]string {
	chars := map[string][]string{}
	for _, s := range strs {
		key := []byte(s)
		sort.Slice(key, func(i, j int) bool {
			return key[i] < key[j]
		})
		chars[string(key)] = append(chars[string(key)], s)
	}
	var res [][]string
	for _, v := range chars {
		res = append(res, v)
	}
	return res
}

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