【LeetCode】746. Min Cost Climbing Stairs（C++）

地址：https://leetcode.com/problems/min-cost-climbing-stairs/

题目：

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是一样的思路，用dp。
其中dp[i]为在位置i的最小的cost。当停留在位置i的时候，可能是从i-2或者i-1上来的。则爬i的最小cost，就是爬i-1的cost和爬i-2的cost里的最小值，加上cost[i]。
其中dp[0]=cost[0]，表明从0出发，dp[1]=cost[1]，从1出发，因为直接走两步从1出发的cost一定小于先到0后到1的。

实现：

class Solution {
public:
	int minCostClimbingStairs(vector& cost) {
		vector 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]);
	}
};

感冒还是没有完全好啊。。困