LeetCode刷题——Array篇(Min Cost Climbing Stairs)

最新推荐文章于 2024-02-11 09:17:00 发布

原创最新推荐文章于 2024-02-11 09:17:00 发布 · 191 阅读

0 ·

CC 4.0 BY-SA版权

Array 专栏收录该内容

16 篇文章

订阅专栏

本文介绍了一个经典的动态规划问题——爬楼梯。目标是最小化到达楼顶的成本，每次可以选择上一阶或两阶。通过动态规划算法，计算出从任意起点出发到达顶层所需的最低成本。

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].

动态规划，寻找最小值，达到第0，1无花费；
达到2：从0跳2步
达到3：从2跳1步
以此类推
这里写图片描述

换成程序来说，
如果我们用一个数组dp[]来存放到达每一层所需要的花费值。则最终结果是求dp[cost.length]的值。因为每次可以走1层或者2层，并且可以从0或者从1开始，所以可以得到dp[0]为0，dp[1]为0。从2开始，dp[i]可以由通过dp[i-2]走2层或者通过dp[i-1]走一层到达，而这i-2和i-1层所要花费的值分别为cost[i-2]和cost[i-1]，所以，dp[i] = min(dp[i-2] + cost[i-2], dp[i-1] + cost[i-1])。该算法的空间复杂度为O(n)，时间复杂度为O(n)。

class Solution {
public:
    int minCostClimbingStairs(vector<int>& cost) {
        int count_size = cost.size()+1;
        int dp0 = 0;
        int dp1 = 0;
        int dp2 = 0;
        for (int i = 2; i < count_size; i++)
        {
            dp2 = min(dp0+cost[i-2],dp1+cost[i-1]);
            dp0 = dp1;
            dp1 = dp2;
            //cout << "到"<<i<<"花费为：" << dp2 << endl;
        }
        return dp2;
    }
};