The Phantom 2016Java Internet Direction One of the questions--frog jumping steps

Problem Description: A frog can jump up to 1 steps at a time, can also jump up to level 2 ... It can also jump up n levels. How many jumps are there in total for the frog to jump up the N-level steps?

Topic Analysis:

1 here the F (n) represents the n steps there is a 1,2,... N-order number of hops.

2 n = 1 o'clock, only 1 kinds of jumps, f (1) = 1

3 N = 2 o'clock, there will be two jumps, a 1-order or 2-order, which returns to the problem (1), f (2) = f (2-1) + f (2-2)

4 n = 3 o'clock, there will be three ways to jump, 1, 2, 3,

So that is the first time to jump out of the 1-step after the rest: F (3-1); first jump out of 2, leaving the F (3-2), the first 3 order, then the rest F (3-3)

So the conclusion is F (3) = f (3-1) +f (3-2) +f (3-3)

5) n = n, there will be N in the way, 1-step, 2-order ... n-order, come to the conclusion:

F (n) = f (n-1) +f (n-2) +...+f (n (n-1)) + f (n-n) => F (0) + F (1) + F (2) + F (3) + ... + f (n-1)

6 The above is already a conclusion, but for simplicity we can continue to simplify:

F (n-1) = f (0) + F (1) +f (2) +f (3) + ... + f ((n-1)-1) = f (0) + F (1) + F (2) + F (3) + ... + f (n-2)

F (n) = f (0) + F (1) + F (2) + F (3) + ... + f (n-2) + f (n-1) = f (n-1) + f (n-1)

Can be drawn:

F (n) = 2*f (n-1)

7 The final conclusion, in the N-step, once there are 1, 2 、... N-order Jump mode, the total jump method is:

| 1, (n=0)

F (n) = | 1, (n=1) | 2*f (n-1), (n>=2)

public class Solutionn {
public static int Jumpfloor (int target) {
if (target <= 0) {
return-1;
} else
if (target = = 1) {
return 1;
} else {
Return 2 * Jumpfloor (TARGET-1);
}
}

public static void Main (string[] args) {
System.out.println (Jumpfloor (3));
}
}

Simple summary:

Because of n steps, the first step has n kinds of jump method: Jump 1, jump 2, to jump N-Class
Jump Level 1, left n-1 level, then the rest of the jump is F (n-1)
Jump Level 2, left n-2 level, then the rest of the jump is F (n-2)
So f (n) =f (n-1) +f (n-2) +...+f (1)
Since f (n-1) =f (n-2) +f (n-3) +...+f (1)
So f (n) =2*f (n-1)