题解 | #跳台阶扩展问题#

f(n) = f(n-1)+f(n-2).....+f(1)+f(0) f(0)=1 f(n-1) = f(n-2)+f(n-3)+f(n-4).....+f(1)+f(0) 上面两个式子相减 f(n)=2*f(n-1)

public class Solution {
    public int jumpFloorII(int target) {
       if(target == 1){
           return 1;
       }
        return 2*jumpFloorII(target-1);
    }
}