A frog can jump up one step at a time, or two steps at a time … it can also jump up n steps. Find out how many jumping methods the frog can use to jump up an n-step.

  java, question

Debugging for half a day tips

java.lang.NoSuchMethodError: main
 Exception in thread "main"

The program I wrote is below. Is there any obvious mistake?

public class Solution {
 
 public int JumpFloor(int target) {
 if (target <= 0) {
 return -1;
 } else if (target == 1) {
 return 1;
 } else {
 return 2 * JumpFloor(target - 1);
 bracket
 bracket
 
 public static void main(String[] args) {
 Solution st = new Solution();
 st.JumpFloor(3);
 bracket
 bracket

In fact, it is the question of Fibonacci number.

Suppose f(n) is the number of n steps.

  1. f(1) = 1

  2. F(2) will have two ways to jump, first order or second order, which returns to the problem f(1),f(2) = f(2-1) plus f(2-2)

  3. F(3) can jump in three ways, first order, second order and third order, then it is left after the first jump out of first order: f(3-1); Jump out of 2nd order for the first time, leaving F (3-2); For the first third order, then f(3-3) remains. So the conclusion is
    F(3) = f(3-1) plus f(3-2) plus f(3-3)

  4. When f(n), there will be a jump in n, order 1, order 2 … order n, and the conclusion is as follows:

F(n) = f(n-1) plus f(n-2) plus … plus f(n-(n-1)) plus f(n-n) => f(0) plus f(1) plus f(2) plus f(3) plus … plus f(n-1) == f(n) = 2*f(n-1)

Therefore, it can be concluded that

Attached code

public long jumpFloor(int n) {
 if (n <= 0)
 return -1;
 if (n == 1)
 return 1;
 return 2 * jumpFloor(n - 1);
 bracket

Considering the efficiency, it can also be done by iteration.