[leetcode] Anagrams

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

Note: All inputs will be in lower-case.

https://oj.leetcode.com/problems/anagrams/

思路：anagram的题目，CC150也有，如何判断是anagram，一种是用hashmap存字符数来判断，另一种是排序后字符相同来判断。这个题目采用排序后的字符作为hashmap的key，是anagram的都归为一组。

import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;

public class Solution {
	public ArrayList<String> anagrams(String[] strs) {
		if (strs == null)
			return null;
		ArrayList<String> res = new ArrayList<String>();
		HashMap<String, ArrayList<String>> group = new HashMap<String, ArrayList<String>>();

		for (String each : strs) {
			char[] tmp = each.toCharArray();
			Arrays.sort(tmp);
			String key = new String(tmp);
			if (group.containsKey(key)) {
				group.get(key).add(each);
			} else {
				ArrayList<String> newList = new ArrayList<String>();
				newList.add(each);
				group.put(key, newList);
			}

		}

		for (String each : group.keySet()) {
			if (group.get(each).size() > 1) {
				res.addAll(group.get(each));
			}
		}

		return res;

	}

	public static void main(String[] args) {
		System.out.println(new Solution().anagrams(new String[] { "abc", "acb",
				"cba", "aaa", "bb", "bb" }));
	}

}

第二遍记录：

public class Solution {
    public List<String> anagrams(String[] strs) {
        List<String> res = new ArrayList<String>();
        if(strs==null||strs.length==0)
            return res;
        
        HashMap<String,List<String>> group = new HashMap<String,List<String>>();
        for(String each:strs){
            char[] charArr = each.toCharArray();
            Arrays.sort(charArr);
            String newStr = new String(charArr);
            if(group.containsKey(newStr)){
                group.get(newStr).add(each);
            }else{
                List<String> tmp = new ArrayList<String>();
                tmp.add(each);
                group.put(newStr,tmp);
            }
            
        }
        
        for(String each:group.keySet()){
            if(group.get(each).size()>1)
                res.addAll(group.get(each));
        }
        return res;
    }
}

参考：

http://blog.csdn.net/linhuanmars/article/details/21664747