#leetcode#500 Keyboard Row

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Given a List of words, return the words that can be typed using letters of alphabet on only one row's of American keyboard like the image below.

Example 1:

Input: ["Hello", "Alaska", "Dad", "Peace"]
Output: ["Alaska", "Dad"]

Note:

1. You may use one character in the keyboard more than once.
2. You may assume the input string will only contain letters of alphabet.

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没有太动脑子想一些简洁的代码， 通过这个题发现自己对  Arrays.asList()， list.toArray(), hashset.addAll() 这几个method用的不熟

public class Solution {
    public String[] findWords(String[] words) {
        if(words == null || words.length == 0)
            return new String[]{};
            
        List<String> res = new ArrayList<>();
        Set<Character> set1 = new HashSet<>();
        Set<Character> set2 = new HashSet<>();
        Set<Character> set3 = new HashSet<>();
        Character[] arr1 = new Character[]{'Q','q','W','w','E','e','R','r','T','t','Y','y','U','u','I','i','O','o','P','p'};
        Character[] arr2 = new Character[]{'A','a','S','s','D','d','F','f','G','g','H','h','J','j','K','k','L','l'};
        Character[] arr3 = new Character[]{'Z','z','X','x','C','c','V','v','B','b','N','n','M','m'};
        set1.addAll(Arrays.asList(arr1));
        set2.addAll(Arrays.asList(arr2));
        set3.addAll(Arrays.asList(arr3));
        
        for(String s : words){
            boolean row1 = false;
            boolean row2 = false;
            boolean row3 = false;
            for(int i = 0; i < s.length(); i++){
                Character c = s.charAt(i);
                if(set1.contains(c))
                    row1 = true;
                else if(set2.contains(c))
                    row2 = true;
                else if(set3.contains(c))
                    row3 = true;
            }
            
            if((row1 && !row2 && ! row3) || (!row1 && row2 && !row3) || (!row1 && !row2 && row3)){
                    res.add(s);
            }
        }
        
        // String[] result = new String[res.size()];    //also works
        // return res.toArray(result);                      
        return res.toArray(new String[0]);
    }
}