Day29 字母异位词分组

给定一个字符串数组，将字母异位词组合在一起。字母异位词指字母相同，但排列不同的字符串

https://leetcode-cn.com/problems/group-anagrams/

示例1：

输入: ["eat", "tea", "tan", "ate", "nat", "bat"]
输出:
[
["ate","eat","tea"],
["nat","tan"],
["bat"]
]

提示：

所有输入均为小写字母。
不考虑答案输出的顺序。

Java解法

思路：

• 因为位置不一致字母一致，考虑用哈希解决，但初步设想用 char值和来确定
• 最后参考官方解解决，排序再比较

package sj.shimmer.algorithm.ten_3;

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

/**
 * Created by SJ on 2021/2/22.
 */

class D29 {
    public static void main(String[] args) {
//        System.out.println(groupAnagrams(new String[]{"eat", "tea", "tan", "ate", "nat", "bat"}));
        System.out.println(groupAnagrams(new String[]{"","b"}));
        System.out.println(groupAnagrams(new String[]{"eat","tea","tan","ate","nat","bat"}));
    }

    public static List<List<String>> groupAnagrams(String[] strs) {
        Map<String, List<String>> map = new HashMap<String, List<String>>();
        for (String str : strs) {
            char[] array = str.toCharArray();
            Arrays.sort(array);
            String key = new String(array);
            List<String> list = map.getOrDefault(key, new ArrayList<String>());
            list.add(str);
            map.put(key, list);
        }
        return new ArrayList<List<String>>(map.values());
    }
}

image

官方解

https://leetcode-cn.com/problems/group-anagrams/solution/zi-mu-yi-wei-ci-fen-zu-by-leetcode-solut-gyoc/

1. 排序

   如上；转换为char数组，排序，做key进行存放

   • 时间复杂度：O(nklogk)
   • 空间复杂度：O(nk)

2. 计数

   用每个字母出现的个数来做标志进行存储，算是变相排序了吧

   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();
               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
               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());
       }
   }