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+1

Recursion

public int result(int n) { if(n==1) return 2; else return 2*result(n-1); } Whatโ€™s is the output of result(5)? Can you please show me the steps as well! ๐Ÿ™๐Ÿผ

4/2/2018 3:52:07 PM

Moe A

3 Answers

New Answer

+5

The first time the function runs this happens n=5;--> return 2*result(4) n=4--> return 2*result(3) therefore. n=5->return 2*2*result(3). n=3--> return 2*result(2) therefore. n=5-> return 2*2*2*result(2) n=2-->return 2*result(1) therefore. n=5-> return 2*2*2*2*result(1) n=1-->return 2 therefore. n=5->return 2*2*2*2*2 =2^5=32. result(5)=32.

+3

the result of result(5) is (2*result (4)) (2*(2*result(3))) (2*(2*(2*result(2)))) (2*(2*(2*(2*result(1))))) (2*(2*(2*(2*(2))))) 32

+2

The secret of recursion is to start counting at the stop condition. It works in most cases. if(n==1) means that your inputs have to be equal to or bigger than 1, so start counting at 1. The next number will always depend on the one you calculated before: n==1 -> 2 n==2 -> 2*2=4 n==3 -> 2*4=8 n==4 -> 2*8=16 n==5 -> 2*16=32 You're basically calculating the powers of 2 here, so you could add this to complete it: if(n==0) return 1;