【LeetCode】746.Min_Cost_Climbing_Stairs

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最新推荐文章于 2024-02-01 07:30:00 发布

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本文介绍了一种使用动态规划求解最小成本爬楼梯问题的方法。给定每级楼梯的成本，目标是以最低成本到达楼顶，可以从第0级或第1级开始。通过计算到达每个台阶的最小总成本来解决问题。

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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].

【解决思路】

动态规划解决。

子问题空间：total_cost记录到达第i个台阶的totalcost（前i个台阶的总cost）；
关联问题变量与子问题：只有两种状态，从（i-1）走一阶到达i，花费为total_cost[i-1]+cost[i-1]；或从（i-2）走两阶到达i，花费为total_cost[i-2]+cost[i-2]。
自底向上计算：只有一个台阶和没有台阶时总花费为0.

【时间复杂度】

遍历一次，耗费时间O(n).

class Solution {
public:
    int minCostClimbingStairs(vector<int>& cost) {
        int cost_i_1=0,cost_i_2=0;
        
        int n=cost.size();
        int total_cost=0;
 
		if(n==1||n==0)
		{
			return 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;
    }
};