剑指offer

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。

code:

分析:
  • f(0) = 1;
  • f(1) = 1;
  • f(2) = f(2-1) + f(2-2) = f(0) + f(1);
  • f(3) = f(3-1) + f(3-2) + f(3-3) = f(0) + f(1) + f(2);
  • f(4) = f(4-1) + f(4-2) + f(4-3) +  f(4-4) = f(0) + f(1) + f(2) + f(3);
  • f(n) = f(n-1) + f(n-2) + f(n-3) + ... + f(n-n) = f(n-1) + f(n-1) = 2 * f(n-1);

递归:

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);
        }
    }
}

动态规划:

public class Solution {
    public int JumpFloorII(int target) {
        if (target <= 0) {
            return 0;
        }
        if (target == 1) {
            return 1;
        }
        int x = 1;
        int y = 2;
        for (int i = 2; i <= target; i++) {
            y = 2 * x;
            x = y;
        }
        return y;
    }
}
原文地址:https://www.cnblogs.com/s-star/p/12512819.html