[leetcode]273. Integer to English Words 整数转英文单词

Posted 2021-01-13 程序媛詹妮弗

Convert a non-negative integer to its english words representation. Given input is guaranteed to be less than 231 - 1.

Example 1:

Input: 123
Output: "One Hundred Twenty Three"
Example 2:

Input: 12345
Output: "Twelve Thousand Three Hundred Forty Five"

 

题目：

给定一个整数，像check一样将该数字用英文读出来

 

思路：

1. Coz given input is less than 2³¹ - 1, we can say that the highest digit level is billion 

2. We divide input into N parts, all the parts follow the similar partten of converting.  We use helper function to help recursive call.

 

代码：

 

 1 class Solution {
 2     // coz array index begins at 0, so we use "" to empty a space
 3     private final String[] belowTen = new String[] {"", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine"};
 4     private final String[] belowTwenty = new String[] {"Ten", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Nineteen"};
 5     private final String[] belowHundred = new String[] {"", "Ten", "Twenty", "Thirty", "Forty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety"};
 6     
 7     public String numberToWords(int num) {
 8         if (num == 0) return "Zero";
 9         return helper(num); 
10     }
11     
12     private String helper(int num) {
13         String result = new String();
14         if (num < 10) result = belowTen[num];
15         else if (num < 20) result = belowTwenty[num -10];
16         // if num is smaller than 100, keep tens digit and add units digit
17         else if (num < 100) result = belowHundred[num/10] + " " + helper(num % 10);
18         // if num is smaller than thousand, call helper to get how many hundreds, call the helper to get the rest digits 
19         else if (num < 1000) result = helper(num/100) + " Hundred " +  helper(num % 100);
20         // if num is smaller than million, call helper to get how many thousands, call the helper to get the rest digits 
21         else if (num < 1000000) result = helper(num/1000) + " Thousand " +  helper(num % 1000);
22         // if num is smaller than billion, call helper to get how many millions, call the helper to get the rest digits 
23         else if (num < 1000000000) result = helper(num/1000000) + " Million " +  helper(num % 1000000);
24         else result = helper(num/1000000000) + " Billion " + helper(num % 1000000000);
25         return result.trim();
26     }
27 }