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 :

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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:

Java Challenge - sumDigits Recursive

Given a non-negative int n, return the sum of its digits recursively (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). sumDigits(126) → 9 sumDigits(49) → 13 sumDigits(12) → 3

Hisoka's Trick :

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public int sumDigits(int n) { if(n < 10) return n; return n%10 + sumDigits(n/10); }

Test Result: