bunnyEars2 challenge

Hi,

I am going through below challenge.

http://codingbat.com/prob/p107330


I have not understood below description

Recursion-1 > bunnyEars2
prev  |  next  |  chance
We have bunnies standing in a line, numbered 1, 2, ... The odd bunnies (1, 3, ..) have the normal 2 ears. The even bunnies (2, 4, ..) we'll say have 3 ears, because they each have a raised foot. Recursively return the number of "ears" in the bunny line 1, 2, ... n (without loops or multiplication).

bunnyEars2(0) → 0
bunnyEars2(1) → 2
bunnyEars2(2) → 5
So even bunnies has 2 ears and odd ones have 3 ears and we have to return total ears?
please advise
LVL 7
gudii9Asked:
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CPColinSenior Java ArchitectCommented:
You should figure out the first challenge before you continue to its sequel. You've essentially just posted ten questions that all boil down to, "I don't understand recursion." Figure out one challenge first, so you get the hang of how recursion works, before you proceed to a pile of others.
awking00Information Technology SpecialistCommented:
In the bunnyEars challenge, I tried to show you how if you keep passing in (bunnies - 1) to your method and adding 2 you eventually will reach the base case. You need to get that method to work, then the modification for this challenge simply becomes adding 2 for an even number of bunnies and 3 for an odd number. Keep in mind that when you pass in (bunnies - 1), you're changing an odd number to an even number and vice versa.

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gudii9Author Commented:
pseudeo code:
1. if o bunnies return ears 0;
2. if one bunny ears 0 i.e 2+bunnyEars(1-1)
3. if n bunnies and odd numbered 2+bunneyEars(n-1)
4. if n bunnies and even numbered 3+bunneyEars(n-1)
5 return ear count
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gudii9Author Commented:
public int bunnyEars2(int bunnies) {
  int result=0;
 if(bunnies==0){
   result=0;
 }
 else{
   if(bunnies%2==1){
   result=2+bunnyEars2(bunnies-1);
   }
    else if(bunnies%2==0){
   result=3+bunnyEars2(bunnies-1);
   }
 }
 return result;
}

Open in new window


above passed all tests. any improvement or alternate approaches?
awking00Information Technology SpecialistCommented:
No need for the result variable -
public int bunnyEars2(int bunnies) {
 if(bunnies==0){
   return 0;
 }
 if(bunnies%2==1) {
   return 2+bunnyEars2(bunnies-1);
 } else  {
   return 3+bunnyEars2(bunnies-1);
 }
}

15:}
gudii9Author Commented:
public int bunnyEars2(int bunnies) {
 if(bunnies==0){
   return 0;
 } 
 if(bunnies%2==1) {
   return 2+bunnyEars2(bunnies-1);
 } else  {
   return 3+bunnyEars2(bunnies-1);
 }
}

Open in new window

i like above approach much better without result
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