Lintcode 1054. Min Cost Climbing Stairs (Easy) (Python)

最新推荐文章于 2021-07-31 11:59:54 发布

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最新推荐文章于 2021-07-31 11:59:54 发布

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本文链接：https://blog.youkuaiyun.com/weixin_41677877/article/details/81039095

本文介绍了一种寻找从楼梯底部到达顶部最小成本的算法。给定每个阶梯的成本，可以选择每次爬一阶或两阶，从第0阶或第1阶开始。通过动态规划方法，记录到达每一阶的最小成本，并最终确定到达顶层的最低消费。

1054. 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].

Code:

class Solution:
    """
    @param cost: an array
    @return: minimum cost to reach the top of the floor
    """
    def minCostClimbingStairs(self, cost):
        # Write your code here
        l = len(cost)
        if (l == 0):
            return 0
        if (l == 1):
            return cost[0]
        if (l == 2):
            return cost.min()
        sto = []
        sto.append(cost[0])
        sto.append(cost[1])
        for i in range(2, l):
            i1 = sto[i-1]
            i2 = sto[i-2]
            m = min(i1,i2)
            sto.append(m+cost[i])
        return min(sto[l-1], sto[l-2])