用文字打印数字 [英] Print number in words

问题描述

我正在尝试使用单词打印三位数的代码。
但如果右边的前两位数介于11和11之间，则无效。 19（包括）。

I am trying code to print a three digit number in words. But this isn't working if first two digits from right are between 11 & 19 (both including).

任何帮助？

package com;

import java.util.Stack;

public class TestMain {
public static void main(String[] args) {

Integer i=512;
int temp =i;int pos=1;
Stack<String> stack=new Stack<String>();
while(temp>0){

    int rem=temp%10;
    temp=temp/10;
    if(rem!=0){stack.push(getString(rem, pos));}
    pos*=10;
}
do{
    System.out.print(stack.pop()+" ");
}while(!stack.isEmpty());



}
static String getString(int i,int position){
    String str=null;

    if(position==10){
        i*=position;
    }

    switch(i){
    case 1:
        str= "One";break;

    case 2:
        str= "two";break;
    case 3:
        str= "three";break;
    case 4:
        str= "four";break;
    case 5:
        str= "five";break;
    case 6:
        str= "six";break;
    case 7:
        str= "seven";break;
    case 8:
        str= "eight";break;
    case 9:
        str= "nine";break;
    case 10:
        str= "Ten";break;
    case 11:
        str= "Eleven";break;
    case 12:
        str= "Twelve";break;
    case 13:
        str= "thirteen";break;
    case 14:
        str= "fourteen";break;
    case 15:
        str= "fifteen";break;
    case 16:
        str= "sixteen";break;
    case 17:
        str= "seventeen";break;
    case 18:
        str= "eighteen";break;
    case 19:
        str= "Nineteen";    break;
    case 20:
        str= "twenty";break;
    case 30:
        str= "Thirty";break;
    case 40:
        str= "forty";break;
    case 50:
        str= "Fifty";break;
    case 60:
        str= "sixty";break;
    case 70:
        str= "Seventy";break;
    case 80:
        str= "Eighty";  break;
    case 90:
        str= "Ninety";break;
    case 100:
        str= "Hundred";
        break;
    }
    if(position>=100){
    str=str+ " "+getString(position, 0);
    }
    return str;
}

}

推荐答案

对于三位数字（负数或正数），您只需要将其分成几个部分，并记住小于20的数字是特殊的（在任何一百个数字中）。作为伪代码（嗯，Python，实际上但足够接近伪代码），我们首先定义查找表：

For a three-digit number (negative or positive), you just need to divide it into sections and remember that numbers less than 20 are special (in any block of a hundred). As pseudo-code (well, Python, actually but close enough to pseudo-code), we first define the lookup tables:

nums1 = ["","one","two","three","four","five","six","seven",
         "eight","nine","ten","eleven","twelve","thirteen",
         "fourteen","fifteen","sixteen","seventeen","eighteen",
         "nineteen"]
nums2 = ["","","twenty","thirty","forty","fifty","sixty",
         "seventy","eighty","ninety"]

请注意，对于10到19之间的数字，没有 onety 。如前所述，每百个块中的二十个以下的数字是专门处理的。

Note that there's no onety for numbers between ten and nineteen. As previously mentioned, numbers under twenty in each block of a hundred are treated specially.

然后我们有一个函数，它首先检查输入值并处理负数作为一个级递归调用。它还处理零的特殊情况：

Then we have the function which first checks the input value and handles negative numbers as a one-level recursive call. It also handles the special case of zero:

def speak (n):
    if n < -999 or n > 999:
        return "out of range"

    if n < 0:
        return "negative " + speak (-n)

    if n == 0:
        return "zero"

下一步是算出数字中的三位数：

Next step is to work out the three digits in the number:

    hundreds = int (n / 100)
    tens = int ((n % 100) / 10)
    ones = n % 10

数百个很简单，因为它只有零到九：

Hundreds are simple since it's only zero through nine:

    if hundreds > 0:
        retstr = nums1[hundreds] + " hundred"
        if tens != 0 or ones != 0:
            retstr = retstr + " and "
    else:
        retstr = ""

其余的只是将零到十九的值视为特殊值，否则我们将其视为 X ty Y （如四十二或七十七）：

The rest is simply treating values from zero to nineteen as special, otherwise we treat it as Xty Y (like forty two or seventy seven):

    if tens != 0 or ones != 0:
        if tens < 2:
            retstr = retstr + nums1[tens*10+ones]
        else:
            retstr = retstr + nums2[tens]
            if ones != 0:
                retstr = retstr + " " + nums1[ones]
    return retstr

快速和底部的脏测试套件：

And a quick and dirty test suite at the bottom:

for i in range (-1000, 1001):
    print "%5d %s"%(i, speak (i))

产生：

-1000 out of range
 -999 negative nine hundred and ninety nine
 -998 negative nine hundred and ninety eight
    : 
   -2 negative two
   -1 negative one
    0 zero
    1 one
    2 two
    :
   10 ten
   11 eleven
   12 twelve
    :
  998 nine hundred and ninety eight
  999 nine hundred and ninety nine
 1000 out of range

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