变态跳台阶（递归 循环）

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。

分析：

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

可以得出：
f(n) = 2*f(n-1)

思路一：递归

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

思路二：递推

public class Solution {    public int JumpFloorII(int target) {        int f = 1;        int fn = 1;        if (target <= 2) return target;        else        {            for (int i = 2; i <= target; i++)            {                fn = 2 * f;                f = fn;            }        }        return fn;    }}