Given an array of strings, group anagrams together.

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

[
  ["ate", "eat","tea"],
  ["nat","tan"],
  ["bat"]
]

一刷
題解：
思路很簡(jiǎn)單，遍歷字符串?dāng)?shù)組，將字符串變?yōu)閏har array, 然后sort, 變成統(tǒng)一的字符串，然后判斷map中是否存在該字符串。這里要注意幾個(gè)字符串和char array互相轉(zhuǎn)換的函數(shù)

Time Complexity - O(n * mlogm) ， Space Complexity - O(m * n)， m為字符串的平均長(zhǎng)度

public class Solution {
    public List<List<String>> groupAnagrams(String[] strs) {
        List<List<String>> res = new ArrayList<>();
        if (strs == null) return res;
        Map<String, List<String>> map = new HashMap<>();
        for(String s:strs){
            char[] sArr = s.toCharArray();
            Arrays.sort(sArr);
            String key = String.valueOf(sArr);
            if(!map.containsKey(key)) map.put(key, new ArrayList<String>());
            map.get(key).add(s);
        }
        
        return new ArrayList<>(map.values());
    }
}