Group Anagrams

Given an array of strings, group anagrams together.

For example, given: ["eat", "tea", "tan", "ate", "nat", "bat"], 
Return:

[
  ["ate", "eat","tea"],
  ["nat","tan"],
  ["bat"]
]
Note:
For the return value, each inner list's elements must follow the lexicographic order.
All inputs will be in lower-case.

Analysis

• anagram经过排序对应一样的字符串
• 用map存储数据, map的key存储排序过的string，map的value存数相对应这个排序字符串的List
• 最后把所有的value输出

Code

public class Solution {
    public List<List<String>> groupAnagrams(String[] strs) {
        List<List<String>> res = new ArrayList<List<String>>();
        if (strs==null || strs.length==0) return res;

        HashMap<String, List<String>> map = new HashMap<String, List<String>>();

        for (String str:strs){
            char[] chars = str.toCharArray();
            Arrays.sort(chars);
            String sorted = new String(chars);
            if (!map.containsKey(sorted)){
                map.put(sorted, new ArrayList<String>(Arrays.asList(str)));
            }
            else {
                //不能写成下面这样的一句，因为list.add() returns boolean
                //map.put(sorted, map.get(sorted).add(str))
                List<String> list = map.get(sorted);
                list.add(str);
                map.put(sorted, list);
            }
        }

        for (String key : map.keySet()){
            //不能写成下面这样的一句，因为collections.sort() returns void
            //res.add(Collections.sort(map.get(key)))
            List<String> list = map.get(key);
            Collections.sort(list);
            res.add(list);
        }

        return res;
    }
}

Note

• 题目不难，思路清楚很容易想出来
• 注意string和char array转换的几个函数
  • str = String.valueOf(charArray) 或者 =new String(strChar);
  • charArray = str.toCharArray()
• map 函数
  • map.put(key, value)
  • value = map.get(key)
  • Set = map.keySet()
  • map.Contains(key)
• sort list
  • Collections.sort(list)