A step has a total of n, if you can jump 1 levels at a time, you can also jump 2 levels, the total number of total hop method, and analyze the algorithm time complexity

Package ms100;/** * A step with a total of n-level, if you can jump 1 levels, you can jump 2 levels, the total number of total hop method, and analyze the algorithm time complexity * Note: This problem has recently appeared, including MicroStrategy More attention to the algorithm of the company has selected a problem as a face test or pen test. First of all, we consider the simplest case: if there is only 1 steps, that obviously there is only one jump method, if there are 2 steps, there are two ways to jump: One is divided into two jumps, each jumping 1 levels, and the other is to jump 2 level. Now let's talk about the general situation: we think of the N-Step jumping method as a function of N, which is recorded as f (n). When N>2, the first jump when there are two different choices: first, only jump 1, when the number of hops is equal to the rest of the n-1 steps of the number of jumps, that is f (n-1); Another option is the first jump 2, at this time the number of hops is equal to the rest of the n-2 step of the number of jumps, That is f (n-2). So the total number of different hops for n-Steps is f (n) = f (n-1) + f (n-2). We summarize the above analysis with a formula as follows:       /  1  (n=1) f (n) =  2  (n=2)       \  F (n-1) + (f-2)  (n>2) */public class ms_27 {private int jumpstep (int n) {if (n<0) return 0;if (n==1| | n==2) return N;return jumpstep (n-1) + jumpstep (n-2);} public static void Main (string[] args) {ms_27 ms27 = new ms_27 (); System.out.println (Ms27.jumpstep (2));}}

A step has a total of n, if you can jump 1 levels at a time, you can also jump 2 levels, the total number of total hop method, and analyze the algorithm time complexity