变态跳台阶

解题思路：有n个台阶时，F(n)=F(n-1)+F(n-2)+F(n-3)+……+F(1)+F(0).又因为F(1)=F(0)=1,经计算可得，F(n)=2^(n-1)

import java.lang.Math;
public class Solution {
    public int JumpFloorII(int target) {
        if(target<2){
            return 1;
        }
        return (int)Math.pow(2,target-1);
    }
}