【leetcode】500. Keyboard Row

原题

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

解析

输入一组字符串，返回所有能在键盘同一行进行输入的字符串

思路

已知键盘的三行字母，用来匹配每个字符串，返回能够全部匹配的字符串

我的解法

    //三行键盘字母
    private static String row1 = "qwertyuiop";
    private static String row2 = "asdfghjkl";
    private static String row3 = "zxcvbnm";

    //主方法
    public static String[] findWords(String[] words) {
        List<String> output = new ArrayList<String>();
        for (int i = 0; i < words.length; i++) {
            char[] chars = words[i].toLowerCase().toCharArray();
            if (row1.contains(String.valueOf(chars[0])) && testRow(row1, chars)) {
                output.add(words[i]);
                continue;
            }
            if (row2.contains(String.valueOf(chars[0])) && testRow(row2, chars)) {
                output.add(words[i]);
                continue;
            }
            if (row3.contains(String.valueOf(chars[0])) && testRow(row3, chars)) {
                output.add(words[i]);
                continue;
            }
        }
        String[] outputArray = new String[output.size()];
        return output.toArray(outputArray);
    }
    //测试该字符串的每一个字符是否在同一行
    private static Boolean testRow(String testString, char[] chars) {
        boolean isAdd = true;
        for (int j = 1; j < chars.length; j++) {
            if (!testString.contains(String.valueOf(chars[j]))) {
                isAdd = false;
                break;
            }
        }
        return isAdd;
    }

最优解法

leetcode上面有人写了一行的最优解。。。。五体投地。。
运用了java8的Stream,和正则表达式，直接匹配一串字符，而不是一个一个匹配

public static String[] findWordsOptimize(String[] words) {
        return Stream.of(words).filter(s -> s.toLowerCase().matches("[qwertyuiop]+|[asdfghjkl]+|[zxcvbnm]+")).toArray(String[]::new);
    }