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?
分析：动态规划，我们可以采取两种方式到达n，一种是从n-1climb 1 step，另一种是从n-2 climb 2 step。时间复杂度O(n), 空间复杂度O(n)。代码如下：
class Solution { public: int climbStairs(int n) { vector<int> record(n+1,0); record[0] = 1; record[1] = 1; for(int i = 2; i <= n; i++){ record[i] = record[i-1] + record[i-2]; } return record[n]; } };
原文：http://www.cnblogs.com/Kai-Xing/p/4141826.html