Climbing Stairs

最新推荐文章于 2024-01-18 13:59:59 发布

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本文探讨了一个经典的编程面试题：爬楼梯问题。通过对递归方法的优化，作者发现该问题实质上是斐波那契数列的应用，并提供了一种高效的循环解决方案。

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?

刚开始看到题目简单的写了个递归，提交超时。后来又仔细推演，发现这不就是斐波那契数列么，于是简单的写个循环提交AC

int climbStairs(int n) 
{
	//if (n < 1) return 0;
	//else if (n == 1) return 1;
	//else if (n == 2) return 2;
	//else return climbStairs(n-1) + climbStairs(n-2);        

	int first = 0, second = 1;
	int fib = first + second;

	for (int i = 2; i <= n; i++)
	{
		first = second;
		second = fib;
		fib = first + second;
	}
	return fib;
}