HackMD
983.Minimum Cost For Tickets
tags: Medium,Array,DP
題目描述
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 for7days:2,3,4,5,6,7, and8.
Return the minimum number of dollars you need to travel every day in the given list of days.
範例
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 daysis in strictly increasing order.costs.length== 3- 1 <=
costs[i]<= 1000
解答
Javascript
思路：
從頭開始遍歷，只要遇到出遊的日子就要買票，選擇1、7、30天中最便宜的更新dp[i]，dp[i]為第i天的最低花費。
function mincostTickets(days, costs) {
const n = days[days.length - 1];
const dp = new Array(n + 1).fill(0);
for (let i = 1; i < n + 1; i++) {
if (days.includes(i)) {
const oneDay = dp[i - 1] + costs[0];
const sevenDay = dp[Math.max(0, i - 7)] + costs[1];
const thirtyDay = dp[Math.max(0, i - 30)] + costs[2];
dp[i] = Math.min(oneDay, sevenDay, thirtyDay);
} else {
dp[i] = dp[i - 1];
}
}
return dp[n];
}
要注意邊界問題，不然就會有神秘的錯誤QQ
MarsgoatMar 28, 2023
Python
class Solution:
def mincostTickets(self, days: List[int], costs: List[int]) -> int:
start, end = days[0], days[-1]
dp = [0] * (end + 1)
for i in range(start, end + 1):
if i in days:
dp[i] = min(dp[i-1] + costs[0], dp[max(0, i-7)] + costs[1], dp[max(0, i-30)] + costs[2])
else:
dp[i] = dp[i-1]
return dp[end]
Ron ChenTue, Mar 28, 2023
