Jump step (perverted version)

problem

A frog can jump on a Class 1 level, you can also hop on level 2 ...... n It can also jump on stage.
The frog jumped seeking a total of n grade level how many jumps.

Algorithm 1

dp问题

• When n = n, there are n hops manner, the first-order, second order ... n order, concluded that:
  F (n) = F (n-1) + F (n-2) + ... + F (n - (n-1)) + f (nn)
• f (n-1) = f (0) + f (1) + f (2) + f (3) + ... + f ((n-1) -1)
  of both can be reduced to l
  f(n)=2f(n-1)

public int JumpFloorII(int target) {

        if(target==1){
            return 1;
        }
        if(target==2){
            return 2;
        }
	//将大问题划分为小问题进行解决
        return JumpFloorII(target-1)*2;


    }

Algorithm 2

Each step has to jump and not jump in both cases (except the last step), the last step must jump. Therefore, the common 2 (n-1) in the case where ^

  //每一个台阶都有跳与不跳两种情况（除了最后一个台阶）最后一个台阶必须跳，所以共有2^(n-1)中情况
    public int JumpFloorII2(int target){
        return  (int)Math.pow(2,target-1);
    }