Problem

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

Notice

All inputs will be in lower-case

Example

Given ["lint", "intl", "inlt", "code"], return ["lint", "inlt", "intl"].

Given ["ab", "ba", "cd", "dc", "e"], return ["ab", "ba", "cd", "dc"].

Note

对变形词的查找和归类，可以将自然排序的词根和所有同根变形词成对存入哈希表map里。然后，返回map.values()里size大于1的字符串数组，再存入结果数组res。注意res并不是ArrayList<ArrayList<String>>()的结构，而是和list数组相同的ArrayList<String>()，所以存list到res一定要用addAll()方法。
有几行容易出错的语句，可以注意一下：
char[] temp = str.toCharArray();
String sortedstr = new String(temp);
for (ArrayList<String> str: map.values()) {};
res.addAll(list);

Solution

public class Solution {
    public List<String> anagrams(String[] strs) {
        List<String> res = new ArrayList<String>();
        HashMap<String, ArrayList<String>> map = new HashMap<String, ArrayList<String>>();
        for (String str: strs) {
            char[] temp = str.toCharArray();
            Arrays.sort(temp);
            String sortedstr = new String(temp);
            if (map.containsKey(sortedstr)) {
                ArrayList<String> list = map.get(sortedstr);
                list.add(str);
                map.put(sortedstr, list);
            }
            else {
                ArrayList<String> list = new ArrayList<String>();
                list.add(str);
                map.put(sortedstr, list);
            }
        }
        for (ArrayList<String> list: map.values()) {
            if (list.size() > 1) res.addAll(list);
        }
        return res;
    }
}