(Leetcode)49. Group Anagrams--Using HashMap

最新推荐文章于 2025-02-03 15:37:10 发布

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最新推荐文章于 2025-02-03 15:37:10 发布

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Problem

49. Group Anagrams

Given an array of strings, group anagrams together.

For example, given: [“eat”, “tea”, “tan”, “ate”, “nat”, “bat”],
Return:

[
[“ate”, “eat”,”tea”],
[“nat”,”tan”],
[“bat”]
]

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Analysis

HashMap

Collection<> values()
返回此映射所包含的值的 Collection 视图。

V put(K key, V value)
参数： key - 指定值将要关联的键；value - 指定键将要关联的值
Return ：与 key 关联的旧值；如果 key 没有任何映射关系，则返回 null。

boolean containsKey(Object key)
Return ：如果此映射包含对于指定键的映射关系，则返回 true。

Solution

java code

package _49GroupAnagrams;

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

public class Solution {
    public List<List<String>> groupAnagrams(String[] strs) {

        HashMap<String, List<String>> markedBoard = new HashMap<String, List<String>>();
        for (int i = 0; i < strs.length; i++) {
            char ch[] = strs[i].toCharArray();
            Arrays.sort(ch);
            String symbolStr = String.valueOf(ch);
            if(!markedBoard.containsKey(symbolStr)){
                List<String> newList = new ArrayList<String>();
                newList.add(strs[i]);
                markedBoard.put(symbolStr, newList);
            }
            else{
                markedBoard.get(symbolStr).add(strs[i]);            
            }
        }
        List<List<String>> Anagrams = new ArrayList<List<String>>(markedBoard.values()); 
        return Anagrams;
    }
    /**
     * @param args
     */
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        String strs[] = {"eat", "tea", "tan", "ate", "nat", "bat"};
        List<List<String>> Anagrams = new Solution().groupAnagrams(strs);
        for (int i = 0; i < Anagrams.size(); i++) {
            for (int j = 0; j < Anagrams.get(i).size(); j++) {
                System.out.print(Anagrams.get(i).get(j)+" ");
            }
            System.out.println();
        }
    }

}