单词打印数字
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19之间（包括两者），则无效。
有帮助吗？
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 stack=new Stack(); 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; } }
对于三位数字（负数或正数），您只需将其划分为多个部分，并记住小于20的数字是特殊的（在任何一百个块中）。 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：
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，10到19之间的数字没有任何onety 。 如前所述，每百个街区中的二十个以下的数字是专门处理的。
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级递归调用。 它还处理零的特殊情况：
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"
下一步是计算出数字中的三位数：
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：
if hundreds > 0: retstr = nums1[hundreds] + " hundred" if tens != 0 or ones != 0: retstr = retstr + " and " else: retstr = ""
其余的只是将从零到十九的值视为特殊值，否则我们将其视为X ty Y （如forty two或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
底部有一个快速而肮脏的测试套件：
for i in range (-1000, 1001): print "%5d %s"%(i, speak (i))
生产：
-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
这是Java中的代码：
private static String[] DIGIT_WORDS = { "zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine" }; private static String[] TENS_WORDS = { "ten", "twenty", "thirty", "forty", "fifty", "sixty", "seventy", "eighty", "ninety" }; private static String[] TEENS_WORDS = { "ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eighteen", "nineteen" }; private static String getHundredWords(int num) { if (num > 999 || num < 0) throw new IllegalArgumentException( "Cannot get hundred word of a number not in the range 0-999"); if (num == 0) return "zero"; String ret = ""; if (num > 99) { ret += DIGIT_WORDS[num / 100] + " hundred "; num %= 100; } if (num < 20 && num > 9) { ret += TEENS_WORDS[num % 10]; } else if (num < 10 && num > 0) { ret += DIGIT_WORDS[num]; } else if (num != 0) { ret += TENS_WORDS[num / 10 - 1]; if (num % 10 != 0) { ret += " " + DIGIT_WORDS[num % 10]; }} return ret; }
- 你必须特别对待零。
- 您在测试主程序中使用模10，因此您在被调用函数中看不到11-20。 实际上，你也不清楚你是否会看到20,30,40等。
- 您的资本化不一致; 优秀的程序员对于一致性是偏执的。