Given an array of strings, return all groups of strings that are anagrams.

Note: All inputs will be in lower-case.

/*
首先搞明白，什么是Anagrams：变换(或颠倒)字母顺序构成的词(或短语)
再就是题目的意思，给出几个字符串，找出其中为Anagrams的字符串
既然是Anagrams，当然至少2个了
Input:　["tea","and","ate","eat","den"]
Output: ["tea","ate","eat"]
*/
public class Solution {
    public List<String> anagrams(String[] strs) {
        List<String> list = new ArrayList<String>();
        Map<String,Integer> map = new HashMap<String,Integer>();
        for(int i = 0;i<strs.length;i++) {
            char[] bytes = strs[i].toCharArray();
            Arrays.sort(bytes);
            String sb = new String(bytes);
            if(map.containsKey(sb)) {
                int index = map.get(sb);
                //不等于-1，则index记录的是第一次,则将其加入list
                if(index != -1) {
                    map.put(sb,-1);
                    list.add(strs[index]);
                }
                list.add(strs[i]);
            } else {
                map.put(sb,i);
            }
        }
        //去掉重复反而错list = new ArrayList<String>(new HashSet<String>(list));
        return list;
    }
}