Cracking the Coding Interview - Fooo

Solutions to Chapter 19 | Moderate
19 6 Given an integer between 0 and 999,999, print an English phrase that describes the
integer (eg, “One Thousand, Two Hundred and Thirty Four”)
SOLUTION
2 7 1
Cracking the Coding Interview | Additional Review Problems
pg 89
This is not an especially challenging problem, but it is a long and tedious one Your interviewer
is unlikely to ask to see every detail, but he / she will be interested in how you approach
the problem
1 public static String numtostring(int num) {
2 StringBuilder sb = new StringBuilder();
3
4 // Count number of digits in num.
5 int len = 1;
6 while (Math.pow((double)10, (double)len ) < num) {
7 len++;
8 }
9
10 String[] wordarr1 = {“”,”One ”, “Two ”, “Three ”, “Four ”,
11 “Five ”, “Six ”, “Seven ”, “Eight ”,”Nine ”};
12 String[] wordarr11 = {“”, “Eleven ”, “Twelve ”, “Thirteen ”,
13 “Fourteen ”, “Fifteen ”, “Sixteen ”,
14 “Seventeen ”, “Eighteen ”, “Nineteen ”};
15 String[] wordarr10 = {“”,”Ten ”, “Twenty ”, “Thirty ”, “Forty ”,
16 “Fifty ”, “Sixty ”, “Seventy ”, “Eighty ”,
17 “Ninety “};
18 String[] wordarr100 = {“”, “Hundred ”, “Thousand ”};
19 int tmp;
20 if (num == 0) {
21 sb.append(“Zero”);
22 } else {
23 if (len > 3 && len % 2 == 0) {
24 len++;
25 }
26 do {
27 // Number greater than 999
28 if (len > 3) {
29 tmp = (num / (int)Math.pow((double)10,(double)len-2));
30 // If tmp is 2 digit number and not a multiple of 10
31 if (tmp / 10 == 1 && tmp%10 != 0) {
32 sb.append(wordarr11[tmp % 10]) ;
33 } else {
34 sb.append(wordarr10[tmp / 10]);
35 sb.append(wordarr1[tmp % 10]);
36 }