[LeetCode] 983. Minimum Cost For Tickets

轉自LeetCode

You have planned some train traveling one year in advance. The days of the year in which you will travel are given as an integer array days. Each day is an integer from 1 to 365.

Train tickets are sold in three different ways:

• a 1-day pass is sold for costs[0] dollars,
• a 7-day pass is sold for costs[1] dollars, and
• 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: 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.

 

Constraints:

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

Solution

Array 求極值，可以從 DP 下手

總共有三種票，日票，週票和月票

定義 dp[i] 為 days[i] 所需要的最少花費

那在推導 dp 的值時，這邊需要兩層 for loop

外層就是每天買日票的最少花費

內層看買週票和買月票的最少花費

日票 -> dp[i] = min(dp[i], dp[i-1] + costs[0])，現在的值，和前一天的最少值加今天買日票之間取最小值

週票 -> dp[i] = min(dp[i], dp[i-1] + costs[1])，現在的值，和買週票前的最少值加今天買週票之間取最小值

月票 -> dp[i] = min(dp[i], dp[i-1] + costs[2])，現在的值，和買月票前的最少值加今天買月票之間取最小值

內層的for迴圈，注意天數的判斷即可

kotlin