LeetCode - Algorithms - 49. Group Anagrams

Problem

49. Group Anagrams

Given an array of strings, group anagrams together.

JavaScript

solution by Joan Indiana Lyness

Algorithms 101: Group Anagrams in JavaScript

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/**
 * @param {string[]} strs
 * @return {string[][]}
 */
var groupAnagrams = function(strs) {
    var m = {};
    for(var i=0;i<strs.length;i++) {
        var s = strs[i];
        var c = s.split("").sort();
        m[c]?m[c].push(s):m[c]=[s];
    }
    var r = [];
    var keys = Object.keys(m);
    for(var k=0; k<keys.length; k++) {
        r.push( m[keys[k]] );
    }
    return r;
};

Submission Detail

  • 101 / 101 test cases passed.
  • Runtime: 144 ms, faster than 32.01% of JavaScript online submissions for Group Anagrams.
  • Memory Usage: 45.6 MB, less than 34.78% of JavaScript online submissions for Group Anagrams.

Java

translation from solution of Joan Indiana Lyness

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import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.Arrays;

class Solution {
    public List<List<String>> groupAnagrams(String[] strs) {
		List<List<String>> list = new ArrayList<List<String>>();
		Map<String, List<String>> map = new HashMap<String, List<String>>();
		String s;
		char[] c;
		String key;
		for (int i = 0; i < strs.length; i++) {
			s = strs[i];
			c = s.toCharArray();
			Arrays.sort(c);
			key = new String(c);
			if (map.containsKey(key)) {
				map.get(key).add(s);
			} else {
				List<String> glist = new ArrayList<String>();
				glist.add(s);
				map.put(key, glist);
			}
		}

		list.addAll(map.values());
		return list;      
    }
}

Submission Detail

  • 101 / 101 test cases passed.
  • Runtime: 9 ms, faster than 83.05% of Java online submissions for Group Anagrams.
  • Memory Usage: 40.8 MB, less than 96.49% of Java online submissions for Group Anagrams.

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