Keyboard Row

题意：

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.

就是让我们判断输入的字符串是否在键盘的同一行上。

思路很简单，先做记录，再看看有没有什么高明的方法。

private static char[][] tables = new char[3][];
    static{
        tables[0] = "QWERTYUIOP".toCharArray();
        tables[1] = "ASDFGHJKL".toCharArray();
        tables[2] = "ZXCVBNM".toCharArray();
    }
    public String[] findWords(String[] words) {
        if(words == null){
            return null;
        }
        List<String> list = new ArrayList<String>();
        for(int i=0; i<words.length; i++){
            String str = words[i];
            if(str == null){
                continue;
            }
            char[] checkChars = str.toUpperCase().toCharArray();
            char[] checkedChars = new char[26];//保存已受检的字符
            //查找目标行
            int target = 0;
            for(;target < tables.length; target++){
                if(contains(tables[target], checkChars[0])){
                    checkedChars[checkChars[0] - 'A'] = 1;
                    break;
                }
            }
            char[] targetChar = tables[target];
            //检查剩下的字符
            boolean isValid = true;
            for(int index = 1; index < checkChars.length; index++){
                char checkChar = checkChars[index];
                if(checkedChars[checkChar - 'A'] != 0){
                    
                }else if(!contains(targetChar, checkChar)){
                    isValid = false;
                    break;
                }
            }
            if(isValid){
                list.add(str);
            }
        }
        return list.toArray(new String[list.size()]);
    }
    
    private boolean contains(char[] chars, char targetChar){
        for(int i=0; i<chars.length; i++){
            if(chars[i] == targetChar)
                return true;
        }
        return false;
    }