# LeetCode 49. Group Anagrams (Java)
###### tags: `leetcode` `Java` `hashmap`

## Description
> Given an array of strings strs, group the anagrams together. You can return the answer in any order.
An Anagram is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.

>給一字符串數組, 將錯位詞(指相同字符不同排列的字符串) 分組

## Example

Input: strs = ["eat","tea","tan","ate","nat","bat"]
Output: [["bat"],["nat","tan"],["ate","eat","tea"]]

Input: strs = [""]
Output: [[""]]

## Idea
- 利用迴圈將string陣列分類進``map``
- 整個程式碼運用到==List==及==Map==的語法(List有兩種實現類：ArrayList,LinkedList)
- [Array和ArrayList的差別](https://ithelp.ithome.com.tw/articles/10229699) [Map用法](https://ithelp.ithome.com.tw/articles/10234424)
- 最後回傳時，要``map``裡的``value``轉換成ArrayList才可回傳

## Code

class Solution {
    public List<List<String>> groupAnagrams(String[] strs) {
        Map<String, ArrayList<String>> map = new HashMap<>();
        
        for (String s: strs){
            char[] key_str = s.toCharArray();//string 轉成char 為了排序
            Arrays.sort(key_str);
            String key = String.valueOf(key_str);//轉回string
            if(map.containsKey(key)){//判斷此key是否在map
               map.get(key).add(s); //有的話將s加入該key的value中
            }
            else{
                map.put(key, new ArrayList<>());//沒有的話將key加入map，並在value建立一個類別
                map.get(key).add(s);
            }
                
        }
        return new ArrayList<>(map.values());
    }
}

## Result
Runtime: **6** ms, faster than **80.28%** of Java online submissions for Group Anagrams.
Memory Usage: **42** MB, less than **81.58%** of Java online submissions for Group Anagrams.