LintCode刷题8

描述

 

把一个非负整数转成英文单词的形式。给定的输入一定比2^31-1小。

 

样例

123 -> "One Hundred Twenty Three"
12345 -> "Twelve Thousand Three Hundred Forty Five"
1234567 -> "One Million Two Hundred Thirty Four Thousand Five Hundred Sixty Seven"

 解题思路，将某个数字分为四个部分，分别为十亿、百万、千、个区分即可，需要注意符号之间的空格

public class Exp1305 {
    /**
     * @param num: a non-negative integer
     * @return: english words representation
     */
    public static String numberToWords(int num) {

        String param =if(num==0){
            return "Zero";
        } "";
        //1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20
        String Param1[] = {"Zero", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine", "Ten", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Nineteen"};
        //30,40,50,60,70,80,90
        String Param2[] = {"Twenty", "Thirty", "Forty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety"};
        if (num > 999999999) {
            param += Param1[(num / 1000000000)];
            if (num % 1000000000 == 0) {
                //如果出现几十亿后面没有任何数字的时候则没有空格
                param += " billion";
                return param;
            }
            //否则加入空格
            param += " Billion ";
        }

        num %= 1000000000;
        int n0_9991 = num / 1000000;
        //999 000 000 - 001 000 000
        if (n0_9991 > 0 && n0_9991 < 1000) {
            if (n0_9991 / 100 > 0) {
                //0-999是需要不加后缀 所以需要区分
                param += Param1[(n0_9991 / 100)];
                param += " Hundred";
                if (n0_9991 % 100 != 0) {
                    param += " ";
                }
            }
            n0_9991 %= 100;
            if (n0_9991 < 20) {
                //1-19
                param += Param1[n0_9991];
            } else {
                //20-99
                param += Param2[(n0_9991 / 10) - 2];
                if (n0_9991 % 10 != 0) {
                    param += " ";
                    param += Param1[n0_9991 % 10];
                }
            }
            param += " Million";
            if(num % 1000000!=0){
                param += " ";
            }
        }


        num %= 1000000;
        int n0_9992 = num / 1000;
        //999 000 - 001 000
        if (n0_9992 > 0 && n0_9992 < 1000) {
            if (n0_9992 / 100 > 0) {
                //0-999是需要不加后缀 所以需要区分
                param += Param1[(n0_9992 / 100)];
                param += " Hundred";
                if (n0_9992 % 100 != 0) {
                    param += " ";
                }
            }
            n0_9992 %= 100;
            if (n0_9992 < 20) {
                //1-19
                param += Param1[n0_9992];
            } else {
                //20-99
                param += Param2[(n0_9992 / 10) - 2];
                if (n0_9992 % 10 != 0) {
                    param += " ";
                    param += Param1[n0_9992 % 10];
                }
            }
            param += " Thousand";
            if(num % 1000!=0){
                param += " ";
            }
        }

        int n0_999 = num %= 1000;
        //0-999
        if (n0_999 > 0 && n0_999 < 1000) {
            if (n0_999 / 100 > 0) {
                //0-999是需要不加后缀 所以需要区分
                param += Param1[(n0_999 / 100)];
                param += " Hundred";
                if (n0_999 % 100 == 0) {
                    //300就不需要后面加空格，然后return 即可
                    param += "";
                    return param;
                } else {
                    param += " ";
                }
            }
            n0_999 %= 100;
            if (n0_999 < 20) {
                //1-19
                param += Param1[n0_999];
            } else {
                //20-99
                param += Param2[(n0_999 / 10) - 2];
                if (n0_999 % 10 != 0) {
                    param += " ";
                    param += Param1[n0_999 % 10];
                }
            }
        }
        return param;
    }

    public static void main(String[] args) {
        //for(int i = 1;i<1000;i++)
        System.out.println(numberToWords(123));
    }
}