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

解法：
這個(gè)也是斐波那契數(shù)列的一個(gè)變種，好多題目都喜歡考這個(gè)，本質(zhì)上是一樣的?？紤]到最后一步可以跨一步或者兩步，那么s(n) = s(n-1) + s(n-1)

class Solution {
public:
    int climbStairs(int n) {
        if(n == 1) {
            return 1;
        }
        
        if(n == 2) {
            return 2;
        }
        
        int curResult = 2;
        int lastResult = 1;
        for(int i=3; i<=n; i++) {
            int temp = curResult;
            curResult = curResult + lastResult;
            lastResult = temp;
        }
        return curResult;
    }
};

依然需要注意一下邊界，最好可以在腦海里面代入具體的值演算一遍。不要犯低級(jí)錯(cuò)誤。