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

Note: All inputs will be in lower-case.

解题思路:首先要理解,什么是anagrams,ie。“tea”、“tae”、“aet”,然后就十分好做了,new一个hashmap,使用一个排过序的String作为key,重复了就往里面添加元素,这里出现一个小插曲,就是排序的时候不要用PriorityQueue,因为PriorityQueue是采用堆排序的,仅保证堆顶元素为优先级最高的(害了我好久)JAVA实现如下:

	static public List<String> anagrams(String[] strs) {
List<String> list = new ArrayList<String>();
HashMap<String, List<String>> hm = new HashMap<String, List<String>>();
for (int i = 0; i < strs.length; i++) {
char [] c=strs[i].toCharArray();
Arrays.sort(c);
String sortString=new String(c);
if (!hm.containsKey(sortString))
hm.put(sortString.toString(), new ArrayList<String>());
hm.get(sortString.toString()).add(strs[i]);
}
Iterator<String> iterator = hm.keySet().iterator();
while (iterator.hasNext()) {
String key = iterator.next();
if (hm.get(key).size() > 1)
list.addAll(hm.get(key));
}
return list;
}

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