package com.lxf.test;

import java.util.*;

public class GroupAnagrams {
    public static void main(String[] args) {
        List<List<String>> lists = groupAnagrams(new String[]{"cab","tin","pew","duh","may","ill","buy","bar","max","doc"});
        for (List<String> list : lists) {
            System.out.println(list);
        }
    }
    public static List<List<String>> groupAnagrams(String[] strs) {
        if (strs==null) {
            return null;
        }
        int length=strs.length;
        //新建存储变位词map
        Map<String,List<String>> map=new HashMap<>();
        for (int i = 0; i < length; i++) {
            //将当前字符串中的字符排序
            char[] chars = strs[i].toCharArray();
            Arrays.sort(chars);
            String key=String.valueOf(chars);
            //排序完字符串如果是变位词肯定一样
            //按key-value（str,List<String>）的形式存储
            if(map.get(key)==null){
                //如果map中没有该key，则新建
                map.put(key,new ArrayList<String>(6));
            }
            map.get(key).add(strs[i]);
        }
        return new ArrayList<>(map.values());
    }
}

2.计数(官方)

class Solution {
    public List<List<String>> groupAnagrams(String[] strs) {
        Map<String, List<String>> map = new HashMap<String, List<String>>();
        for (String str : strs) {
            int[] counts = new int[26];
            int length = str.length();
            //记录下字符串所有字符出现的次数
            for (int i = 0; i < length; i++) {
                counts[str.charAt(i) - 'a']++;
            }
            // 将每个出现次数大于 0 的字母和出现次数按顺序拼接成字符串，作为哈希表的键
            // 实际上就是按考虑字符顺序，从a-z按次数拼接成map的key
            StringBuffer sb = new StringBuffer();
            for (int i = 0; i < 26; i++) {
                if (counts[i] != 0) {
                    sb.append((char) ('a' + i));
                    sb.append(counts[i]);
                }
            }
            String key = sb.toString();
            List<String> list = map.getOrDefault(key, new ArrayList<String>());
            list.add(str);
            map.put(key, list);
        }
        return new ArrayList<List<String>>(map.values());
    }
}