[Sword Point offer] abnormal jumping steps

Title Description

A frog can jump up to 1 steps at a time, or jump up to level 2 ... It can also jump on n levels. To ask the frog to jump on an n-level stair total number of hops

Enter a description

Number of steps

Output description

Jump Method Number

Problem analysis

• Set N-order hop number to F (n)

• When N=1, f (1) = 1

• When the n=2, divided into the last step 2 and jump 1 order two cases, there is f (2) =f (0) +f (1) =1+1=2

• When n=3, divided into the last step to jump 3 order, jump 2 order and jump 1 order three cases, there is f (3) =f (0) +f (1) +f (2) =1+1+2=4

• With f (n) = f (n-1) +f (n-2) +...+f (1) + F (0) established
  At the same time F (n-1) =f (n-2) +...+f (1) + f (0) is established, can be obtained f (n) =2f (n-1) (n>=2)

　

It is obvious that a recursive formula can be derived:

| 1 (n=0)
F (n) = | 1 (n=1)
| 2*f (n-1) (n>=2)

Solution one run time: 35ms occupied memory: 654k

publicclass Solution {    publicintJumpFloorII(int target) {        if(target<=1)  return1;        return2*JumpFloorII(target-1);    }}

Solution two run time: 34ms occupied memory: 654k

publicclass Solution {    publicintJumpFloorII(int target) {   if(target<=1)  return1;        return1<<(target-1);    }}

Another way to think about it: a total of n steps, the last step is bound to jump up, the other n-1 can jump not jump, altogether how many total situation?

2 (n-1)

Here, multiply with the shift, faster in time!

[Sword Point offer] abnormal jumping steps