如何将字母数字电话号码转换为数字 [英] How to convert an alphanumeric phone number to digits

问题描述

更新:

我的实用程序的最终版本如下所示:

The final version of my utility looks like this:

StringBuilder b = new StringBuilder();

for(char c : inLetters.toLowerCase().toCharArray())
{
    switch(c)
    {
    case '0':                                          b.append("0"); break;
    case '1':                                          b.append("1"); break;
    case '2': case 'a': case 'b': case 'c':            b.append("2"); break;
    case '3': case 'd': case 'e': case 'f':            b.append("3"); break;
    case '4': case 'g': case 'h': case 'i':            b.append("4"); break;
    case '5': case 'j': case 'k': case 'l':            b.append("5"); break;
    case '6': case 'm': case 'n': case 'o':            b.append("6"); break;
    case '7': case 'p': case 'q': case 'r': case 's':  b.append("7"); break;
    case '8': case 't': case 'u': case 'v':            b.append("8"); break;
    case '9': case 'w': case 'x': case 'y': case 'z':  b.append("9"); break;
    }
}

return builder.toString();

<小时>

原始问题:

我正在承担将字母数字电话号码转换为一串数字的简单任务.例如，1-800-HI-HAXOR 将变为 1-800-44-42967.我最初的尝试是创建一个讨厌的 switch 语句，但我想要一个更优雅、更有效的解决方案.这是我得到的:

I'm taking on the simple task of converting an alphanumeric phone number to a string of digits. For example, 1-800-HI-HAXOR would become 1-800-44-42967. My initial attempt was to create a nasty switch statement, but I'd love a more elegant, and efficient solution. Here's what I've got:

for(char c : inLetters.toLowerCase().toCharArray())
{
    switch(c)
    {
    case '0':                                         result+="0"; break;
    case '1':                                         result+="1"; break;
    case '2': case 'a': case 'b': case 'c':           result+="2"; break;
    case '3': case 'd': case 'e': case 'f':           result+="3"; break;
    case '4': case 'g': case 'h': case 'i':           result+="4"; break;
    case '5': case 'j': case 'k': case 'l':           result+="5"; break;
    case '6': case 'm': case 'n': case 'o':           result+="6"; break;
    case '7': case 'p': case 'q': case 'r': case 's': result+="7"; break;
    case '8': case 't': case 'u': case 'v':           result+="8"; break;
    case '9': case 'w': case 'x': case 'y': case 'z': result+="9"; break;
    }
}

谢谢！

推荐答案

switch 语句并没有那么糟糕.您的算法与电话号码的长度成线性关系.该代码是可读的，并且很容易通过检查来验证.我不会弄乱它，除了添加一个 default 案例来处理错误.(我不是 Java 程序员，所以如果它被称为别的东西，请原谅我.)

The switch statement is not really that bad. Your algorithm is linear with respect to the length of the phone number. The code is readable and pretty easy to verify by inspection. I wouldn't mess with it, except to add a default case for handling errors. (I'm not a Java programmer, so forgive me if it's called something else.)

如果您有使其更快，则按字符索引的预初始化表将避免超出基本错误检查的任何比较.您甚至可以通过复制表中的值来避免大小写转换 (digit['A'] = digit['a'] = "2";).初始化表的成本将在转换总数中摊销.

If you have to make it faster, a pre-initialized table indexed by character would avoid any comparisons beyond basic error checking. You could even avoid the case conversion by duplicating the values in the table (digit['A'] = digit['a'] = "2";). The cost of initializing the table would be amortized over the total number of conversions.

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