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]]:
        d = defaultdict(list)
        for s in strs:
            k = ''.join(sorted(s))
            d[k].append(s)
        return list(d.values())

class Solution:
    def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
        d = defaultdict(list)
        for s in strs:
            cnt = [0] * 26
            for c in s:
                cnt[ord(c) - ord('a')] += 1
            d[tuple(cnt)].append(s)
        return list(d.values())

Java

class Solution {
    public List<List<String>> groupAnagrams(String[] strs) {
        Map<String, List<String>> d = new HashMap<>();
        for (String s : strs) {
            char[] t = s.toCharArray();
            Arrays.sort(t);
            String k = String.valueOf(t);
            d.computeIfAbsent(k, key -> new ArrayList<>()).add(s);
        }
        return new ArrayList<>(d.values());
    }
}

class Solution {
    public List<List<String>> groupAnagrams(String[] strs) {
        Map<String, List<String>> d = new HashMap<>();
        for (String s : strs) {
            int[] cnt = new int[26];
            for (int i = 0; i < s.length(); ++i) {
                ++cnt[s.charAt(i) - 'a'];
            }
            StringBuilder sb = new StringBuilder();
            for (int i = 0; i < 26; ++i) {
                if (cnt[i] > 0) {
                    sb.append((char) ('a' + i)).append(cnt[i]);
                }
            }
            String k = sb.toString();
            d.computeIfAbsent(k, key -> new ArrayList<>()).add(s);
        }
        return new ArrayList<>(d.values());
    }
}

C++

class Solution {
public:
    vector<vector<string>> groupAnagrams(vector<string>& strs) {
        unordered_map<string, vector<string>> d;
        for (auto& s : strs) {
            string k = s;
            sort(k.begin(), k.end());
            d[k].emplace_back(s);
        }
        vector<vector<string>> ans;
        for (auto& [_, v] : d) ans.emplace_back(v);
        return ans;
    }
};

class Solution {
public:
    vector<vector<string>> groupAnagrams(vector<string>& strs) {
        unordered_map<string, vector<string>> d;
        for (auto& s : strs) {
            int cnt[26] = {0};
            for (auto& c : s) ++cnt[c - 'a'];
            string k;
            for (int i = 0; i < 26; ++i) {
                if (cnt[i]) {
                    k += 'a' + i;
                    k += to_string(cnt[i]);
                }
            }
            d[k].emplace_back(s);
        }
        vector<vector<string>> ans;
        for (auto& [_, v] : d) ans.emplace_back(v);
        return ans;
    }
};

Go

func groupAnagrams(strs []string) (ans [][]string) {
	d := map[string][]string{}
	for _, s := range strs {
		t := []byte(s)
		sort.Slice(t, func(i, j int) bool { return t[i] < t[j] })
		k := string(t)
		d[k] = append(d[k], s)
	}
	for _, v := range d {
		ans = append(ans, v)
	}
	return
}

func groupAnagrams(strs []string) (ans [][]string) {
	d := map[[26]int][]string{}
	for _, s := range strs {
		cnt := [26]int{}
		for _, c := range s {
			cnt[c-'a']++
		}
		d[cnt] = append(d[cnt], s)
	}
	for _, v := range d {
		ans = append(ans, v)
	}
	return
}

TypeScript

function groupAnagrams(strs: string[]): string[][] {
    const d: Map<string, string[]> = new Map();
    for (const s of strs) {
        const k = s.split('').sort().join('');
        if (!d.has(k)) {
            d.set(k, []);
        }
        d.get(k)!.push(s);
    }
    return Array.from(d.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()];
}

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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