30-Day Leetcoding Challenge Day6

本题有两种思路：

第一种是将key-value中的key通过排序存储到set集合，value添加该字符串

第二种是将key-value中的key通过数组计数方式存储，value添加该字符串

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
            for c in s:
                count[ord(c) - ord('a')] += 1 #注意ord函数返回ASCII值
            res[tuple(count)].append(s) #key必须为tuple不可变类型
        return res.values()