变态跳台阶

题目描述

一只青蛙一次可以跳上1级台阶，也可以跳上2级……它也可以跳上n级。求该青蛙跳上一个n级的台阶总共有多少种跳法。

参数的target是台阶的数量:

f(1) = 1

f(2) = f(2-1) + f(2-2)        

f(3) = f(3-1) + f(3-2) + f(3-3) ,第一次跳一阶，后面剩下f(3-1)； 第一次跳两阶，后面剩下f(3-2)；第一次跳三阶，后面剩下f(3-3)

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

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(n-1) + f(n-1) = 2*f(n-1)

             | 1       ,(n=0 ) 

f(n) =     | 1       ,(n=1 )

              | 2*f(n-1),(n>=2)

 

public class Solution {

    public int JumpFloorII(int target) {

        if (target <= 0)

            return 0;

        else if (target == 1)

            return 1;

        else 

            return 2 * JumpFloorII(target - 1);

    }

}

转载于:https://www.cnblogs.com/MarkLeeBYR/p/9777209.html