LeetCode 70. Climbing Stairs

Description:

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?

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

分析：

如果能够看出这是斐波那契数列，解这道题就很容易了。
但是一开始我想着要用动态规划来求解，一开始想着从0步开始，每一次加1或者加2，如下图：

然后我用二维数组来处理，自己想复杂了，一直Wrong Answer，后来灵光一现，这不就是斐波那契数列吗？动态规划一下子就变得简单了。

代码如下：

动态规划：

#include <iostream>
using namespace std;
class Solution {
public:
	int climbStairs(int n) {
		int d[n + 10];
		d[1] = 1;
		d[2] = 2;
		for (int i = 3; i < n + 1; i++) {
			d[i] = d[i - 1] + d[i - 2];
		}
		return d[n];
	}
};
int main() {
	int n;
	cin >> n;
	Solution s;
	cout << s.climbStairs(n) << endl;
	return 0;
}

Fibonacci数列：

#include <iostream>
using namespace std;
class Solution {
public:
	int climbStairs(int n) {
		if (n == 1) return 1;
		int first = 1;
		int second = 2;
		for (int i = 3; i <= n; i++) {
			int third = first + second;
			first = second;
			second = third;
		}
		return second;
	}
};
int main() {
	int n;
	cin >> n;
	Solution s;
	cout << s.climbStairs(n) << endl;
	return 0;
}