[leetcode] 746. Min Cost Climbing Stairs @ python

最新推荐文章于 2025-05-06 11:02:52 发布

原创最新推荐文章于 2025-05-06 11:02:52 发布 · 178 阅读

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博客围绕楼梯最小花费问题展开，给定每级楼梯的非负花费，可从第0或1级起步，每次能爬1或2级。采用动态规划解法，构造数组dp，确定状态转移方程，dp[0]和dp[1]初始为0，dp长度比cost大1，最终返回dp[-1]，时间复杂度O(n)，空间复杂度O(n)。

原题

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].

解法

动态规划. 构造数组dp, dp[i]表示到达i点的最少花费, 转态转移方程:

dp[i] = min(dp[i-2]+cost[i-2], dp[i-1]+cost[i-1])

由于我们可以从index 0或1起步, 因此dp[0] = dp[1] = 0, 最后要到台阶顶部, 因此dp长度需要比cost长度大1, 最后返回dp[-1]即可.
Time: O(n)
Space: O(n)

代码

class Solution(object):
    def minCostClimbingStairs(self, cost):
        """
        :type cost: List[int]
        :rtype: int
        """
        # base case
        if len(cost) <= 2:
            return min(cost)
        dp = [0]*(len(cost)+1)
        
        for i in range(2, len(dp)):
            dp[i] = min(dp[i-2]+cost[i-2], dp[i-1]+cost[i-1])
        return dp[-1]