LeetCode 746 使用最小花费爬楼梯

const int N = 1000;
class Solution {
public:
    int dp[N];
    int minCostClimbingStairs(vector<int>& cost) {
        dp[0] = cost[0];
        dp[1] = cost[1];

        for (int i = 2; i < cost.size(); i ++) //最后一步相当于不花费
            dp[i] = min(dp[i - 1], dp[i - 2]) + cost[i];

        return min(dp[cost.size() - 1], dp[cost.size() - 2]); //最后要么从倒数第二层爬要么从倒数第一层爬
    }
};