Minimum Cost For Tickets (M)

題目

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

題意

旅遊票有三種：包1/7/30天，求遊覽完所有指定天數需要的最少花費。

思路

動態規劃。dp[i]表示遊覽完從第1天到第i天中所有指定天數所需要的最少花費。那麼對於dp[i]有幾種情況：

1. 第i天是不需要遊覽的，那麼dp[i]=dp[i-1];

2. 第i天需要遊覽，再分為3種情況：

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

   取其中的最小值。

程式碼實現

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]];
    }
}