[leetcode]70. 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: 2Output:  2Explanation:  There are two ways to climb to the top.1. 1 step + 1 step2. 2 steps

Example 2:Input: 3Output:  3Explanation:  There are three ways to climb to the top.1. 1 step + 1 step + 1 step2. 1 step + 2 steps3. 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;    }};