leetcode -- climbing stairs

最新推荐文章于 2025-08-11 22:22:04 发布

weixin_33967071

最新推荐文章于 2025-08-11 22:22:04 发布

阅读量107

点赞数

CC 4.0 BY-SA版权

文章标签： python

原文链接：https://my.oschina.net/u/1447519/blog/1591857

本文探讨了一个经典的编程面试题——爬楼梯问题。该问题要求计算出达到楼梯顶部的不同方式的数量。通过递推公式的方式，文章提供了一种高效的解决方案，并指出这个问题实际上可以归结为斐波那契数列的应用。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

2019独角兽企业重金招聘Python工程师标准>>>

Climbing Stairs

You are climbing a stair case. It takes n steps to reach to the top.

Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?

Note: Given n will be a positive integer.

Example 1:

Input: 2
Output:  2

Explanation: There are two ways to climb to the top.

1 step + 1 step
2 steps Example 2:

Input: 3
Output:  3

Explanation: There are three ways to climb to the top.

1 step + 1 step + 1 step
1 step + 2 steps
2 steps + 1 step

class Solution {
    public int climbStairs(int n) {
         if (n <=0 )
            return n;

        if ( 1 == n || 2 == n){
            return n;
        }

        int step1 = 1;
        int step2 = 2;
        int ans = 0;
        for(int i=3; i<=n; i++){
            ans = step1 + step2;
            step1 = step2;
            step2 = ans;
        }

        return ans;
    }
}

its actually a fibonacci!

转载于:https://my.oschina.net/u/1447519/blog/1591857