Description
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.
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] = 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 <= 3651 <= days[i] <= 365daysis in strictly increasing order.costs.length == 31 <= costs[i] <= 1000
Code
DP
Time Complexity: , Space Complexity:
遞迴的邏輯是：在 Day i，假設我們買了一日票，那 total cost 就會是一日票的價錢再加上一天之後的總花費，若買了七日票， total cost 就會是七日票的價錢和七日之後的總花費，三十日的票同理。
class Solution {
public:
int mincostTickets(vector<int>& days, vector<int>& costs) {
unordered_set<int> t(days.begin(), days.end());
int dp[366];
dp[0] = 0;
for(int i = 1; i < 366; i++) {
dp[i] = dp[i - 1];
if(t.find(i) != t.end()) {
dp[i] = min({dp[max(0, i - 1)] + costs[0], dp[max(0, i - 7)] + costs[1], dp[max(0, i - 30)] + costs[2]});
}
}
return dp[365];
}
};