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# 171.Anagrams

## 1.Description(Medium)

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"]`.

[**Challenge**](http://www.lintcode.com/en/problem/anagrams/#challenge)

What is Anagram?

* Two strings are anagram if they can be the same after change the order of characters.

## 2.Code

用一个HashMap\<String,List\<String>>来记录，取出每个string排序然后把与原顺序的string记录进hashmap的list中,最后判断ma。p.values()里面的每个list的size如果>1,就说明有他的anagrams，直接addAll进去就行了。

```
public List<String> anagrams(String[] strs) {
        List<String> result=new ArrayList<String>();
        if(strs==null || strs.length==0){
            return result;
        }

        HashMap<String,List<String>> map=new HashMap<>();
        for(int i=0;i<strs.length;i++){
            //fetch this word and make it array then sort it then remake a string
            char[] current=strs[i].toCharArray();
            Arrays.sort(current);
            String key=new String(current);

            if(map.containsKey(key)){
                map.get(key).add(strs[i]);
            }else{
                List<String> word=new ArrayList<String>();
                word.add(strs[i]);
                map.put(key, word);
            }       
        }


        for(List<String> element:map.values()){
            if(element.size()>1){
                result.addAll(element);
            }
        }
        return result;
    }
```