[Leetcode] 746. Min Cost Climbing Stairs 解题报告_leetcode 746 报告-优快云博客

本文链接：https://blog.youkuaiyun.com/magicbean2/article/details/79526671

本文介绍了一个经典的动态规划问题——如何以最小的成本爬到楼梯顶部。通过定义状态转移方程dp[i]=cost[i]+min(dp[i-2],dp[i-1])来求解，并提供了完整的C++代码实现。

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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].

思路：

这是最简单和最原始的动态规划问题了。我们定义dp[i]表示从第i个位置上开始出发需要付出的最小代价，那么递推公式为：dp[i] = cost[i] + min(dp[i - 2], dp[i - 1])。最后我们返回dp数组中倒数两个元素中的小者即可。

代码：

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