772 · Group Anagrams - LintCode

# Description

To give you a string array, please group the misplaced words(refers to the same character string with different permutations).

All input are lower case

# Example

Example 1:

Input:
["eat","tea","tan","ate","nat","bat"]
Output:
[["ate","eat","tea"],
 ["bat"],
 ["nat","tan"]]

Example 2:

Input:
["eat","nowhere"]
Output:
[["eat"],
 ["nowhere"]]

相同的是：還是一樣要把字符串排序放在 hashMap 裡
不同的是：這次把所有的字符都分組，把 anagram 放在一起，不同的字符都分開放。

要注意的是怎麼分組...

	public List<List<String>> groupAnagram(String[] strs) {
	List<List<String>> res = new ArrayList<List<String>>();
	Map<String, Integer> hm = new HashMap<>();
	int index = 0; //res 的位置
	for (String s : strs) {
	char[] charArray = s.toCharArray();
	Arrays.sort(charArray);
	String sortedStr = String.valueOf(charArray);
	if (!hm.containsKey(sortedStr)) {
	hm.put(sortedStr, index++);
	res.add(new ArrayList<String>());
	}
	res.get(hm.get(sortedStr)).add(s);
	}
	return res;
	}

	hm.putIfAbsent(sortedArr, new ArrayList<>());
	hm.get(sortedArr).add(s);

	return new Arraylist<>(hm.values());

	public List<List<String>> groupAnagrams(String[] strs) {
	Map<String, List<String>> hm = new HashMap<>();
	for (String s: strs) {
	char[] charArr = s.toCharArray();
	Arrays.sort(charArr);
	String sortedArr = String.valueOf(charArr);
	hm.putIfAbsent(sortedArr, new ArrayList<>());
	hm.get(sortedArr).add(s);
	}
	return new Arraylist<>(hm.values());

	}