30-Day Leetcoding Challenge Day6

This question has two ideas:

The first is the key in the key-value stored in the set by ordering the set, value is added to the string

The second is the key in the key-value storage array by counting mode, value added to the string

JAVA

class Solution {
    public List<List<String>> groupAnagrams(String[] strs) {
        if(strs.length == 0)
            return new ArrayList();
        Map<String, List> res = new HashMap<String, List>();
        for(String s : strs){
            char[] ca = s.toCharArray();
            Arrays.sort(ca);
            String key = String.valueOf(ca);
            if(!res.containsKey(key))
                res.put(key, new ArrayList());
            res.get(key).add(s);
        }
        return new ArrayList(res.values());
    }
}

class Solution {
    public List<List<String>> groupAnagrams(String[] strs) {
        if(strs.length == 0)
            return new ArrayList();
        Map<String, List> res = new HashMap<String, List>();
        for(String s : strs){
            char[] ca = s.toCharArray();
            int[] count = new int[26];
            for(int i = 0; i < ca.length; i++){
                count[ca[i] - 'a']++;
            }
            StringBuilder sb = new StringBuilder("");
            for(int i = 0; i < count.length; i++){
                sb.append('#');
                sb.append(count[i]);
            }
            String key = sb.toString();
            if(!res.containsKey(key))
                res.put(key, new ArrayList());
            res.get(key).add(s);
        }
        return new ArrayList(res.values());
    }
}

 

 

Python3

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

class Solution:
     DEF groupAnagrams (Self, STRs: List [STR]) -> List [List [STR]]: 
        RES = collections.defaultdict (List)
         for S in STRs: 
            COUNT = [0] * 26 is
             for C in S: 
                COUNT [ord (C) - ord ( ' A ' )] +. 1 = # that ord function returns the ASCII value of 
            RES [tuple (COUNT)] the append (S). # Key must be immutable type tuple 
        return res.values ()