编程开发习题Climbing Stairs

Question :

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

Solution :

很久之前就做过了，对于爬楼梯，有一步或两步的走法，则有递推式 f(n) = f(n-1) + f(n-2)，f(n) 表示第 n 个楼梯的走法，由在第 n-1 个楼梯走一步 + 在第 n-2 个楼梯走两步得到。

其实就是斐波那契数列。

class Solution {
public:
    int climbStairs(int n) {
        if (n == 1) return 1;
        if (n == 2) return 2;
        int a = 1, b = 2, t;
        for (int i = 2; i < n; i++) {
            t = a + b;
            a = b;
            b = t;
        }
        return t;
    }
};