題目：

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:

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

理解：

和前面的climbing stairs是一樣的思路，用dp。 其中dp[i]為在位置i的最小的cost。當停留在位置i的時候，可能是從i-2或者i-1上來的。則爬i的最小cost，就是爬i-1的cost和爬i-2的cost裡的最小值，加上cost[i]。 其中dp[0]=cost[0]，表明從0出發，dp[1]=cost[1]，從1出發，因為直接走兩步從1出發的cost一定小於先到0後到1的。

實現：

class Solution {
public:
	int minCostClimbingStairs(vector<int>& cost) {
		vector<int> dp(cost.size(), 0);
		dp[0] = cost[0], dp[1] = cost[1];
		for (int i = 2; i < cost.size(); ++i) {
			dp[i] = cost[i] + min(dp[i - 1], dp[i - 2]);
		}
		return min(dp[cost.size() - 1], dp[cost.size() - 2]);
	}
};
 

感冒還是沒有完全好啊。。困