如果两个字符是键盘上的邻居，如何有效地检查？

Question

我想为Android开发一个软键盘，并且已经有一个自动更正算法，如果输入字符和字典中单词的字符是键盘上的邻居，则会根据这个事实提出建议。这与levenshtein算法结合使用（如果必须用不同的字符替换字符，则检查它们是否是邻居）。这就是为什么经常调用此检查的原因。目前，它消耗了50％的自动更正时间。

我目前的做法是一个单独的三层，三层。第一层：第一个字符。第二层：第二个字符：第三层：如果字符是邻居，则保存信息的布尔值。但是我担心特里尼有点矫枉过正？每个孩子的实习生哈希图也会降低它的速度？我应该用自己的charToNumber函数构建一个hashmap吗？

你会怎么做？哪些瓶颈可以避免？每次执行检查时，Character.toLowerCase（）似乎都是低效的。

我希望你能帮助我加快完成任务：）

Answer 1

您只想确定键盘上是否有两个字符相邻？为什么不使用从字符到一组相邻字符的地图？使用高效的数据结构时，您将得到O(1)时间 - 使用数组作为映射（连续密钥空间 - 密钥的ASCII代码），使用BitSet作为一组相邻密钥。也非常紧凑。

以下是示例代码：

BitSet[] adjacentKeys = new BitSet[127];

//initialize
adjacentKeys[(int) 'q'] = new BitSet(127);
adjacentKeys[(int) 'q'].set((int)'a');
adjacentKeys[(int) 'q'].set((int)'w');
adjacentKeys[(int) 'q'].set((int)'s');
//...

//usage
adjacentKeys[(int) 'q'].get((int) 'a');     //q close to a yields true
adjacentKeys[(int) 'q'].get((int) 'e');     //q close to e yields false

这应该非常高效，没有循环和复杂的计算，如hashCode s。当然你必须手动初始化表，我建议在应用程序启动时从som外部配置文件中执行一次。

BTW整洁的想法！

Answer 2

我真的很喜欢这个主意。

对于原始速度，您将使用大量switch语句。代码会很大，但没有什么比这更快了：

public static boolean isNeighbour(char key1, char key2) {
    switch (key1) {
    case 'a':
        return key2 == 'w' || key2 == 'e' || key2 == 'd' || key2 == 'x' || key2 == 'z';
    case 'd':
        return key2 == 's' || key2 == 'w' || key2 == 'f' || key2 == 'c' || key2 == 'x';
    // etc
    default:
        return false;
    }
}

这是一种“标准”方式，它应该仍然表现良好：

private static final Map<Character, List<Character>> neighbours =
    new HashMap<Character, List<Character>>() {{
    put('s', Arrays.asList('a', 'w', 'e', 'd', 'x', 'z')); 
    put('d', Arrays.asList('s', 'e', 'w', 'f', 'c', 'x'));
    // etc
}};

public static boolean isNeighbour(char key1, char key2) {
    List<Character> list = neighbours.get(key1);
    return list != null && list.contains(key2);
}

此算法未使用a isneighbour b然后b isneighbour a，而是为了简化代码而牺牲数据大小这一事实。

Answer 3

如何为每个键分配数字并使用它来确定接近度。

    public static void main(String[] args) {
    double[] d = new double[26];
    d['q'-97] = 100d;
    d['w'-97] = 101d;
    d['e'-97] = 102d;
    d['r'-97] = 103d;
    d['t'-97] = 104d;
    //(optionally, put a space of 5 between right hand and left hand for each row)
    d['y'-97] = 105d;
    d['u'-97] = 106d;
    d['i'-97] = 107d;
    d['o'-97] = 108d;
    d['p'-97] = 109d;


    //my keyboard middle row is about 20% indented from first row
    d['a'-97] = 200.2;
    d['s'-97] = 201.2;
    d['d'-97] = 202.2;
    d['f'-97] = 203.2;
    d['g'-97] = 204.2;
    d['h'-97] = 205.2;
    d['j'-97] = 206.2;
    d['k'-97] = 207.2;
    d['l'-97] = 208.2;

    //third row is about 50% indented from middle row
    d['z'-97] = 300.5;
    d['x'-97] = 301.5;
    d['c'-97] = 302.5;
    d['v'-97] = 303.5;
    d['b'-97] = 304.5;
    d['n'-97] = 305.5;
    d['m'-97] = 306.5;

    for (char a = 'a'; a <= 'z'; a++) {
        for (char b = 'a'; b <= 'z'; b++)
            if (a != b && prox(a,b,d))
                System.out.println(a + " and " + b + " are prox");
    }

}

static boolean prox(char a, char b, double m) {
    double a1 = m[a-97];
    double a2 = m[b-97];

    double d = Math.abs(a1-a2);
    //TODO: add in d == 5 if there is a spacing for left and right hand gap (since it's more unlikely of a crossover)
    return d == 0 || d == 1 || (d >= 99 && d <= 101);
}

部分输出：

a and q are prox
a and s are prox
a and w are prox
a and z are prox
....
g and b are prox
g and f are prox
g and h are prox
g and t are prox
g and v are prox
g and y are prox   
....
y and g are prox
y and h are prox
y and t are prox
y and u are prox 

Answer 4

这是我的匈牙利语版本（如果有人需要的话）：

 public static boolean isHungarianNeighbour(int key1, int key2) {
    switch (key1) {
        case 'q':
            return key2 == 'w' || key2 == 's' || key2 == 'a' || key2 == '1' || key2 == '2';
        case 'w':
            return key2 == 'q' || key2 == '2' || key2 == '3' || key2 == 'e' || key2 == 's' || key2 == 'a';
        case 'e':
            return key2 == '3' || key2 == '4' || key2 == 'w' || key2 == 'r' || key2 == 's' || key2 == 'd';
        case 'r':
            return key2 == '4' || key2 == '5' || key2 == 'e' || key2 == 't' || key2 == 'd'|| key2 == 'f';
        case 't':
            return key2 == '5' || key2 == '6' || key2 == 'r' || key2 == 'z' || key2 == 'f' || key2 == 'g';
        case 'z':
            return key2 == '6' || key2 == '7' || key2 == 't' || key2 == 'u' || key2 == 'g' || key2 == 'h';
        case 'u':
            return key2 == '7' || key2 == '8' || key2 == 'z' || key2 == 'i' || key2 == 'h' || key2 == 'j';
        case 'i':
            return key2 == '8' || key2 == '9' || key2 == 'u' || key2 == 'o' || key2 == 'j' || key2 == 'k';
        case 'o':
            return key2 == '9' || key2 == 'ö' || key2 == 'i' || key2 == 'p' || key2 == 'k' || key2 == 'l';
        case 'p':
            return key2 == 'ö' || key2 == 'ü' || key2 == 'o' || key2 == 'ő' || key2 == 'l' || key2 == 'é';
        case 'ő':
            return key2 == 'ü' || key2 == 'ó' || key2 == 'p' || key2 == 'ú' || key2 == 'é' || key2 == 'á';
        case 'ú':
            return key2 == 'ó' || key2 == 'ő' || key2 == 'á' || key2 == 'ű';
        case 'a':
            return key2 == 'q' || key2 == 'w' || key2 == 's' || key2 == 'y' || key2 == 'í';
        case 's':
            return key2 == 'w' || key2 == 'e' || key2 == 'a' || key2 == 'd' || key2 == 'y' || key2 == 'x';
        case 'd':
            return key2 == 'e' || key2 == 'r' || key2 == 's' || key2 == 'f' || key2 == 'x' || key2 == 'c';
        case 'f':
            return key2 == 'r' || key2 == 't' || key2 == 'd' || key2 == 'g' || key2 == 'c' || key2 == 'v';
        case 'g':
            return key2 == 't' || key2 == 'z' || key2 == 'f' || key2 == 'h' || key2 == 'v' || key2 == 'b';
        case 'h':
            return key2 == 'z' || key2 == 'u' || key2 == 'g' || key2 == 'j' || key2 == 'b' || key2 == 'n';
        case 'j':
            return key2 == 'u' || key2 == 'i' || key2 == 'h' || key2 == 'k' || key2 == 'n' || key2 == 'm';
        case 'k':
            return key2 == 'i' || key2 == 'o' || key2 == 'j' || key2 == 'l' || key2 == 'm';
        case 'l':
            return key2 == 'o' || key2 == 'p' || key2 == 'k' || key2 == 'é';
        case 'é':
            return key2 == 'p' || key2 == 'ő' || key2 == 'l' || key2 == 'á';
        case 'á':
            return key2 == 'ő' || key2 == 'ú' || key2 == 'é' || key2 == 'ű';
        case 'ű':
            return key2 == 'á' || key2 == 'ú';
        case 'í':
            return key2 == 'a' || key2 == 'y';
        case 'y':
            return key2 == 'a' || key2 == 's' || key2 == 'í' || key2 == 'x';
        case 'x':
            return key2 == 's' || key2 == 'd' || key2 == 'y' || key2 == 'c';
        case 'c':
            return key2 == 'd' || key2 == 'f' || key2 == 'x' || key2 == 'v';
        case 'v':
            return key2 == 'f' || key2 == 'g' || key2 == 'c' || key2 == 'b';
        case 'b':
            return key2 == 'g' || key2 == 'h' || key2 == 'v' || key2 == 'n';
        case 'n':
            return key2 == 'h' || key2 == 'j' || key2 == 'b' || key2 == 'm';
        case 'm':
            return key2 == 'j' || key2 == 'k' || key2 == 'n' || key2 == '?';
        default:
            return false;
    }
}