#Leetcode# 746. Min Cost Climbing Stairs

最新推荐文章于 2025-12-22 20:10:38 发布

转载最新推荐文章于 2025-12-22 20:10:38 发布 · 83 阅读

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原文链接：http://www.cnblogs.com/zlrrrr/p/10821715.html

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#数据结构与算法

博客围绕爬楼梯问题展开，在楼梯上每步有非负成本，支付成本后可爬一步或两步，需找出到达楼顶的最小成本，可从索引 0 或 1 的台阶开始，还给出相关代码示例及问题范围说明。

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

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].

代码：

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

每次觉得 dp 没那么难的时候就要明白是错觉！

跑步去啦要减肥呢！

转载于:https://www.cnblogs.com/zlrrrr/p/10821715.html