0983. Minimum Cost For Tickets (M)

Minimum Cost For Tickets (M)

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:

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. 1 <= days.length <= 365
2. 1 <= days[i] <= 365
3. days is in strictly increasing order.
4. costs.length == 3
5. 1 <= costs[i] <= 1000

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题意

旅游票有三种：包1/7/30天，求游览完所有指定天数需要的最少花费。

思路

动态规划。dp[i]表示游览完从第1天到第i天中所有指定天数所需要的最少花费。那么对于dp[i]有几种情况：

1. 第i天是不需要游览的，那么dp[i]=dp[i-1];

2. 第i天需要游览，再分为3种情况：

   1. 在第i天购入1日票，dp[i]=dp[i-1]+costs[0]
   2. 在第i-7天购入7日票，dp[i]=dp[i-7]+costs[1]
   3. 在第i-30天购入30日票，dp[i]=dp[i-30]+costs[2]

   取其中的最小值。

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代码实现

Java

class Solution {
    public int mincostTickets(int[] days, int[] costs) {
        int[] dp = new int[366];
        for (int i = 0; i < days.length; i++) {
            dp[days[i]] = -1;
        }
        for (int i = 1; i < 366; i++) {
            if (dp[i] == 0) {
                dp[i] = dp[i - 1];
            } else {
                int one = dp[i - 1] + costs[0];
                int seven = i > 7 ? dp[i - 7] + costs[1] : costs[1];
                int thirty = i > 30 ? dp[i - 30] + costs[2] : costs[2];
                dp[i] = Math.min(one, Math.min(seven, thirty));
            }
        }
        return dp[days[days.length - 1]];
    }
}