LeetCode 49. Group Anagrams 找相同的字母组成的字符串

题目:

Given an array of strings, group anagrams together.

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

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

Note: All inputs will be in lower-case.

题目解释：相同字母组成的字符串分到一类，其中也要求字母个数相同，用HashMap来判别

public class Solution {
    public List> groupAnagrams(String[] strs) {
        if(strs==null||strs.length==0) return null;
		List> result = new ArrayList>();
        HashMap> store = new HashMap>();
		for(String s : strs){
        	char[] mid = s.toCharArray();
        	Arrays.sort(mid);
        	String new_s = String.valueOf(mid);
        	if(store.containsKey(new_s)){
        		store.get(new_s).add(s);
        	}else{
        		List slist = new ArrayList();
        		slist.add(s);
        		store.put(new_s, slist);
        	}
        }
		Set keyset = store.keySet();
		for(String middle : keyset)
			result.add(store.get(middle));
		return result;
    }
}

注意点：

1.char[] 转成String 要用String.valueof()函数来转化，如果直接用toString()转换成地址，与预想结果不同

2.Java中Set是一个不包含重复元素的集合,添加相同元素则直接被替换掉，访问直接for遍历即可