746. Min Cost Climbing Stairs

最新推荐文章于 2025-01-01 21:30:00 发布

bwcxljsm

最新推荐文章于 2025-01-01 21:30:00 发布

阅读量289

点赞数

CC 4.0 BY-SA版权

分类专栏： leetcode 算法 Leetcode 文章标签： leedcode

本文链接：https://blog.youkuaiyun.com/qq_36718317/article/details/79258157

Leetcode 同时被 3 个专栏收录

89 篇文章

订阅专栏

leetcode

85 篇文章

订阅专栏

算法

83 篇文章

订阅专栏

本文介绍了一种通过动态规划求解在给定每级楼梯成本的情况下，从底部到达顶部楼梯所需的最低成本的方法。以数组形式给出每级楼梯的成本，允许每次移动一步或两步。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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].

题目大意：走第i步花费cost[i]，每步可以走1或2，起点可以是0或1，问走到最后的cost最小是多少。

思路：

典型的动态规划。有一个数组a来保存走到i的最少花费。那么a[i]=min(a[i-1],a[i-2])+cost[i]。最后结果便是min(a[cost.size()-1],a[cost.size()-2]).

代码：

class Solution {
public:
    int minCostClimbingStairs(vector<int>& cost) {
        int a[1001];
        a[0]=cost[0],a[1]=cost[1];
        for(int i=2;i<cost.size();i++)
        {
            a[i]=min(a[i-1],a[i-2])+cost[i];
        }
        return min(a[cost.size()-1],a[cost.size()-2]);
    }
};