[Swift]LeetCode983. 最低票價 | Minimum Cost For Tickets

原文地址：http://www.javashuo.com/article/p-xtcynxoy-md.html 

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

---

在一個火車旅行很受歡迎的國度，你提早一年計劃了一些火車旅行。在接下來的一年裏，你要旅行的日子將以一個名爲 days 的數組給出。每一項是一個從 1 到 365 的整數。

火車票有三種不一樣的銷售方式：

• 一張爲期一天的通行證售價爲 costs[0] 美圓；
• 一張爲期七天的通行證售價爲 costs[1] 美圓；
• 一張爲期三十天的通行證售價爲 costs[2] 美圓。

通行證容許數天無限制的旅行。 例如，若是咱們在第 2 天得到一張爲期 7 天的通行證，那麼咱們能夠連着旅行 7 天：第 2 天、第 3 天、第 4 天、第 5 天、第 6 天、第 7 天和第 8 天。

返回你想要完成在給定的列表 days 中列出的每一天的旅行所須要的最低消費。 

示例 1：

輸入：days = [1,4,6,7,8,20], costs = [2,7,15]
輸出：11
解釋： 
例如，這裏有一種購買通行證的方法，能夠讓你完成你的旅行計劃：
在第 1 天，你花了 costs[0] = $2 買了一張爲期 1 天的通行證，它將在第 1 天生效。
在第 3 天，你花了 costs[1] = $7 買了一張爲期 7 天的通行證，它將在第 3, 4, ..., 9 天生效。
在第 20 天，你花了 costs[0] = $2 買了一張爲期 1 天的通行證，它將在第 20 天生效。
你總共花了 $11，並完成了你計劃的每一天旅行。

示例 2：

輸入：days = [1,2,3,4,5,6,7,8,9,10,30,31], costs = [2,7,15]
輸出：17
解釋：
例如，這裏有一種購買通行證的方法，能夠讓你完成你的旅行計劃： 
在第 1 天，你花了 costs[2] = $15 買了一張爲期 30 天的通行證，它將在第 1, 2, ..., 30 天生效。
在第 31 天，你花了 costs[0] = $2 買了一張爲期 1 天的通行證，它將在第 31 天生效。 
你總共花了 $17，並完成了你計劃的每一天旅行。 

提示：

1. 1 <= days.length <= 365
2. 1 <= days[i] <= 365
3. days 按順序嚴格遞增
4. costs.length == 3
5. 1 <= costs[i] <= 1000

---

124ms

 1 class Solution {
 2     func mincostTickets(_ days: [Int], _ costs: [Int]) -> Int {
 3         var n:Int = days.count
 4         var dp:[Int] = [Int](repeating:Int.max / 2,count:n+1)
 5         dp[0] = 0
 6         for i in 1...n
 7         {
 8             dp[i] = dp[i-1] + costs[0]
 9             for j in (0...(i - 1)).reversed()
10             {
11                 if days[i-1] - days[j] + 1 <= 7
12                 {
13                     dp[i] = min(dp[i], dp[j] + costs[1])
14                 }
15                 if days[i-1] - days[j] + 1 <= 30
16                 {
17                     dp[i] = min(dp[i], dp[j] + costs[2])
18                 }
19             }
20         }
21         return dp[n]
22     }
23 }