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 lowercase English letters.

Solutions

Python3

class Solution:
    def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
        chars = 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);
            chars.computeIfAbsent(k, key -> new ArrayList<>()).add(s);
        }
        return new ArrayList<>(chars.values());
    }
}

TypeScript

function groupAnagrams(strs: string[]): string[][] {
    let map = new Map();
    for (let str of strs) {
        let arr = str.split('');
        arr.sort();
        let key = arr.join('');
        let value = map.get(key) ? map.get(key) : [];
        value.push(str);
        map.set(key, value);
    }
    return Array.from(map.values());
}

function groupAnagrams(strs: string[]): string[][] {
    const map = new Map<string, string[]>();
    for (const str of strs) {
        const k = str.split('').sort().join('');
        map.set(k, (map.get(k) ?? []).concat([str]));
    }
    return [...map.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
}

Rust

use std::collections::HashMap;

impl Solution {
    pub fn group_anagrams(strs: Vec<String>) -> Vec<Vec<String>> {
        let mut map = HashMap::new();
        for s in strs {
            let key = {
                let mut arr = s.chars().collect::<Vec<char>>();
                arr.sort();
                arr.iter().collect::<String>()
            };
            let val = map.entry(key).or_insert(vec![]);
            val.push(s);
        }
        map.into_iter().map(|(_, v)| v).collect()
    }
}

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