On a staircase, the i-th step has some non-negative cost cost[i] assigned (0 indexed).

Once you pay the cost, you can either climb one or two steps. You need to find minimum cost to reach the top of the floor, and you can either start from the step with index 0, or the step with index 1.

Example 1:

Input: cost = [10, 15, 20]
Output: 15
Explanation: Cheapest is start on cost[1], pay that cost and go to the top.

Example 2:

Input: cost = [1, 100, 1, 1, 1, 100, 1, 1, 100, 1]
Output: 6
Explanation: Cheapest is start on cost[0], and only step on 1s, skipping cost[3].

Note:

1. cost will have a length in the range [2, 1000].
2. Every cost[i] will be an integer in the range [0, 999].

這題是相對比較簡單的動態規劃的題目了，可以得到這麽個式子，dp[i] = min(dp[i-1], dp[i-2]) + cost[i], 這題比較要註意的就是最後出要加個0：

 1 class Solution {
 2 public:
 3     int minCostClimbingStairs(vector<int>& cost) {
 4  
 5         cost.push_back(0);//!!
 6         int n = cost.size();
 7         
 8         vector<int> res(n, 0);
 9         res[0] = cost[0];
10         res[1] = cost[1];
11         
12         for(int
 
 i = 2; i < n; ++i)
13         {
14             res[i]=min(res[i-1], res[i-2])+cost[i];
15         }
16         
17         return res[n-1];
18     }
19 };

Min Cost Climbing Stairs