原题链接在这里：https://leetcode.com/problems/minimum-cost-for-tickets/
题目：
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 <= days.length <= 3651 <= days[i] <= 365daysis in strictly increasing order.costs.length == 31 <= costs[i] <= 1000
题解：
For all the days when travel is not needed, its min cost should be the same as previous day.
When it comes to the day travel is needed, there are 3 options, 1 day pass + up to previous day minimum cost, 7 days pass + up to previous 7 days minimum cost, 30 days pass + up to previous 30 days minimum cost.
Time Complexity: O(n). n = days[days.length-1].
Space: O(n).
AC Java:
1 class Solution {
2 public int mincostTickets(int[] days, int[] costs) {
3 int lastDay = days[days.length-1];
4 int [] dp = new int[lastDay+1];
5 int ind = 0;
6 for(int i = days[0]; i<=lastDay; i++){
7 if(i != days[ind]){
8 dp[i] = dp[i-1];
9 }else{
10 dp[i] = Math.min(dp[i-1]+costs[0], Math.min(dp[Math.max(0, i-7)]+costs[1], dp[Math.max(0, i-30)]+costs[2]));
11 ind++;
12 }
13 }
14
15 return dp[lastDay];
16 }
17 }
The truth is we only need to maintain last 30s data.
Thus it couls limit dp array to 30 days.
Time Complexity: O(n).
Space: O(1).
AC Java:
1 class Solution {
2 public int mincostTickets(int[] days, int[] costs) {
3 int [] dp = new int[30];
4 int ind = 0;
5 for(int i = days[0]; i<=days[days.length-1]; i++){
6 if(i != days[ind]){
7 dp[i%30] = dp[(i-1)%30];
8 }else{
9 dp[i%30] = Math.min(dp[(i-1)%30] + costs[0], Math.min(dp[Math.max(0, i-7) % 30] + costs[1], dp[Math.max(0, i-30) % 30] + costs[2]));
10 ind++;
11 }
12 }
13
14 return dp[days[days.length-1]%30];
15 }
16 }