Monday, June 6, 2016

Java Challenge - count7 Recursive

Given a non-negative int n, return the count of the occurrences of 7 as a digit, so for example 717 yields 2. (no loops). Note that mod (%) by 10 yields the rightmost digit (126 % 10 is 6), while divide (/) by 10 removes the rightmost digit (126 / 10 is 12).

count7(717) → 2
count7(7) → 1
count7(123) → 0

Hisoka :
1
2
3
4
5
public int count7(int n) {
  if(n<10 && n == 7) return 1;
  else if(n<10 && n!=7) return 0;
  else return count7(n%10)+count7(n/10);
}

Test Result of Codingbat:

No comments:

Post a Comment