剑指offer、牛客-变态跳台阶

数归法：可得:

public class Solution {
    public int JumpFloorII(int n) {
        if(n <= 0)
            return 0;
        if(n == 1)
            return 1;
        int result = 1;
        for(int k = 2; k <= n; k++){
            result = result * 2;
        }
        return result;
    }
}