500. Keyboard Row

题目：

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.

链接：https://leetcode.com/problems/keyboard-row/#/description

3/29/2017

performance 8ms, beat 15%

注意的问题：

1. 第3，4，5行的初始化

2. 第11，12行的建立set。为什么不能采取1中的方法？Character[]和char[]是不能自动转化的。我对Java了解不清，需要仔细研究。

3. 第6，15行，ArrayList和[]转化

4. set1.retainAll(set2)是求交集，set1.containsAll(set2)是求s2是否为s1的子集

public class Solution {
    public String[] findWords(String[] words) {
        Set<Character> s1 = new HashSet<Character>(Arrays.asList('q', 'w', 'e', 'r', 't', 'y', 'u', 'i', 'o', 'p'));
        Set<Character> s2 = new HashSet<Character>(Arrays.asList('a', 's', 'd', 'f', 'g', 'h', 'j', 'k', 'l'));
        Set<Character> s3 = new HashSet<Character>(Arrays.asList('z', 'x', 'c', 'v', 'b', 'n', 'm'));
        ArrayList<String> tmp = new ArrayList<String>();

        for (int i = 0; i < words.length; i++) {
            Set<Character> s = new HashSet<Character>();
            String t = words[i].toLowerCase();
            for (int j = 0; j < words[i].length(); j++) 
                s.add(t.charAt(j));
            if (s1.containsAll(s) || s2.containsAll(s) || s3.containsAll(s)) tmp.add(words[i]);
        }
        String[] ret = tmp.toArray(new String[tmp.size()]);
        return ret;
    }
}

看别人的算法，基本上如果performance好的话，在循环体里不用containsAll而是自己单独每个character比较。

更多讨论：

https://discuss.leetcode.com/category/649/keyboard-row