[leetcode] 983. Minimum Cost For Tickets @ python

最新推荐文章于 2022-04-06 10:55:14 发布

原创最新推荐文章于 2022-04-06 10:55:14 发布 · 586 阅读

0 ·

CC 4.0 BY-SA版权

博客围绕火车旅行购票问题展开，给定旅行日期数组和三种不同时长车票价格，要求计算最小花费。采用动态规划方法，设dp[i]为到第i天的最少花费，若第i天不出行，dp[i] = dp[i - 1]，并给出示例及代码思路。

原题

In a country popular for train travel, you have planned some train travelling one year in advance. The days of the year that you will travel is given as an array days. Each day is an integer from 1 to 365.

Train tickets are sold in 3 different ways:

a 1-day pass is sold for costs[0] dollars;
a 7-day pass is sold for costs[1] dollars;
a 30-day pass is sold for costs[2] dollars.
The passes allow that many days of consecutive travel. For example, if we get a 7-day pass on day 2, then we can travel for 7 days: day 2, 3, 4, 5, 6, 7, and 8.

Return the minimum number of dollars you need to travel every day in the given list of days.

Example 1:

Input: days = [1,4,6,7,8,20], costs = [2,7,15]
Output: 11
Explanation:
For example, here is one way to buy passes that lets you travel your travel plan:
On day 1, you bought a 1-day pass for costs[0] = $2, which covered day 1.
On day 3, you bought a 7-day pass for costs[1] = $7, which covered days 3, 4, …, 9.
On day 20, you bought a 1-day pass for costs[0] = $2, which covered day 20.
In total you spent $11 and covered all the days of your travel.
Example 2:

Input: days = [1,2,3,4,5,6,7,8,9,10,30,31], costs = [2,7,15]
Output: 17
Explanation:
For example, here is one way to buy passes that lets you travel your travel plan:
On day 1, you bought a 30-day pass for costs[2] = $15 which covered days 1, 2, …, 30.
On day 31, you bought a 1-day pass for costs[0] = $2 which covered day 31.
In total you spent $17 and covered all the days of your travel.

Note:

1 <= days.length <= 365
1 <= days[i] <= 365
days is in strictly increasing order.
costs.length == 3
1 <= costs[i] <= 1000

解法

动态规划, 设dp[i]为到第i天为止的最少花费. 初始化数组dp为0, 动态转移方程:

dp[i] = min(dp[i-1] + costs[0], dp[i-7] + costs[1], dp[i-30] + costs[2])

遍历数组dp, 如果第i天不在days里, 表示当天不出行, 那么dp[i] = dp[i-1]. 为防止index取负数, 这里使用max(i-1, 0)使得index的最小值为0.

代码

class Solution(object):
    def mincostTickets(self, days, costs):
        """
        :type days: List[int]
        :type costs: List[int]
        :rtype: int
        """
        # dp[i] represent the min cost at index i
        dp = [0]*(days[-1]+1)
        for i in range(1, len(dp)):
            if i not in days:
                dp[i] = dp[i-1]
                continue
                
            dp[i] = min(dp[max(0,i-1)] + costs[0], dp[max(0,i-7)] + costs[1], dp[max(0,i-30)] + costs[2])
            
        return dp[-1]