算法分析与设计week13--746. Min Cost Climbing Stairs_min cost climbing stairs descriptionhintssubmissio-优快云博客

本文链接：https://blog.youkuaiyun.com/chenxy258/article/details/79049206

本文介绍了一个经典的动态规划问题——最低成本攀登楼梯。题目要求找出从楼梯底部到达顶部所需的最小花费，可以一步上一个台阶或两个台阶。通过动态规划求解，给出了一种高效的算法实现。

746. Min Cost Climbing Stairs

Description

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

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].

Analysis

题意：最低成本攀登楼梯

思路：使用动态规划，递推公式：dp[i] = cost[i] + min(dp[i - 1], dp[i - 2]);

class Solution {
public:
    int minCostClimbingStairs(vector<int>& cost) {
        int n = cost.size();
        int total_cost = 0;

        if (n == 1 || n == 0) return 0;

        int cost_i_1 = 0, cost_i_2 = 0;

        for (int i = 2; i <= n; i++) {
            total_cost = min(cost[i - 1] + cost_i_1,cost[i - 2] + cost_i_2);
            cost_i_2 = cost_i_1;
            cost_i_1 = total_cost;
        }
        return total_cost;
    }
};