A number "n" is given, the task is to print spelling for n using recursion.
Example 1: Print spelling for 1000
Input: 1000
Output: One Zero Zero Zero
Example 2: Print spelling for 1
Input: 1
Output: One
Example 3: Print spelling for 895
Input: 895
Output: Eight Nine Five
Solutions
Method 1: Recursion
If we take modulo(%) of a number by 10 we get the last digit of that number i.e. 123 % 10 = 3; similarly if we divide a number by 10 and store the value in int we get the number except last digit i.e. int r = 123 / 10 = 12.
We can solve this problem easily using recursion, first we need to maintain an array to get spelling corresponding to a particular number. Now for every n we will take the last digit and get the corresponding spelling from the array and concat this at the end of f(n-1) recursive call result.
If n==0, that means the number is completely processed; return blank string ("") - base condition.
package com.cb.recursion;
/**
* Recursion
*/
public class R6_PrintNumberSpelling {
static String[] mappingArray = {"Zero", "One", "Two", "Three", "Four",
"Five", "Six", "Seven", "Eight", "Nine"};
public static String printSpelling(int n) {
if (n < 1)
return "";
int current = n % 10;
int remaining = n / 10;
return printSpelling(remaining) + " " + mappingArray[current];
}
public static void main(String[] args) {
System.out.println(printSpelling(1000));
System.out.println(printSpelling(1));
System.out.println(printSpelling(126));
System.out.println(printSpelling(895));
}
}
One Zero Zero Zero One One Two Six Eight Nine Five