Climbing Stairs

最新推荐文章于 2021-11-04 20:26:38 发布

原创最新推荐文章于 2021-11-04 20:26:38 发布 · 182 阅读

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本文探讨了一个经典的算法问题——爬楼梯。通过递推的方法，文章详细解释了如何计算达到楼梯顶部的不同方式的数量，并给出了简洁的C++实现代码。

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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. 1 step + 1 step
2. 2 steps

Example 2:

Input: 3
Output:  3
Explanation:  There are three ways to climb to the top.

1. 1 step + 1 step + 1 step
2. 1 step + 2 steps
3. 2 steps + 1 step

class Solution {
public:
    int climbStairs(int n) {
        
        
        if(n == 1)
            return 1;
        else if(n == 2)
            return 2;
        else if(n == 3)
            return 3;
        
        int sum[n];
        sum[0] = 1;
        sum[1] = 2;
        sum[2] = 3;
        
        for(int i=3;i<n;i++)
            sum[i] = sum[i-1]+sum[i-2];
        
        return sum[n-1];
        
        
    }
};

一开始迷了，没想到这么简单，