Java Challenge - fibonacci recursive

The fibonacci sequence is a famous bit of mathematics, and it happens to have a recursive definition. The first two values in the sequence are 0 and 1 (essentially 2 base cases). Each subsequent value is the sum of the previous two values, so the whole sequence is: 0, 1, 1, 2, 3, 5, 8, 13, 21 and so on. Define a recursive fibonacci(n) method that returns the nth fibonacci number, with n=0 representing the start of the sequence. fibonacci(0) → 0 fibonacci(1) → 1 fibonacci(2) → 1

Hisoka :

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public int fibonacci(int n) { if(n == 0 || n == 1) return n; return fibonacci(n-1)+fibonacci(n-2); }

Hasil:

Codingbat:
Codingbat g' ngasih jawaban... :v

Java Challenge - factorial

Given n of 1 or more, return the factorial of n, which is n * (n-1) * (n-2) ... 1. Compute the result recursively (without loops). factorial(1) → 1 factorial(2) → 2 factorial(3) → 6

Hisoka:

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public int factorial(int n) { if(n == 1) return 1; return n*factorial(n-1); }

Hasil:

Codingbat:

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public int factorial(int n) { // Base case: if n is 1, we can return the answer directly if (n == 1) return 1; // Recursion: otherwise make a recursive call with n-1 // (towards the base case), i.e. call factorial(n-1). // Assume the recursive call works correctly, and fix up // what it returns to make our result. return n * factorial(n-1); }

Woooh...sama yaaak... :D