LeetCode 983. Minimum Cost For Tickets

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题目

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:

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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:

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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

思路

DP.

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dp[i]表示旅行到i天为止需要的最少旅行价格。
那么，如果当第i天不需要旅行（不在days中），则这一天就不用额外买票，所以需要花费的价格等于昨天的价格；
如果当第i天需要旅行的话，那么需要求三种买票方式的最小价格：昨天的最少价格+一天的票 costs[0]，7天前的最少价格+7天的票 costs[1] ，30天前的最少价格+30天的票 costs[2]。

则递推公式为：

  • dp[i] = dp[i - 1] 当第i天不用旅行
  • dp[i] = min(dp[i - 1] + costs[0], dp[i - 7] + costs[1], dp[i - 30] + costs[2]) 当第i天需要旅行

注意一下向前查找的时候是否越界。

代码

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class Solution:
    def mincostTickets(self, days: List[int], costs: List[int]) -> int:
        dp = [float("inf")] * 366
        for day in days:
            dp[day] = 0
        dp[0] = 0
        for i in range(1, 366):
            if dp[i] == float("inf"):
                dp[i] = dp[i-1]
            else:
                dp[i] = min(dp[i-1]+costs[0], dp[max(i-7, 0)]+costs[1], dp[max(i-30, 0)]+costs[2])
        return dp[days[-1]]