1235. Maximum Profit in Job Scheduling

We have n jobs, where every job is scheduled to be done from startTime[i] to endTime[i], obtaining a profit of profit[i].

You're given the startTimeendTime and profit arrays, return the maximum profit you can take such that there are no two jobs in the subset with overlapping time range.

If you choose a job that ends at time X you will be able to start another job that starts at time X.

 

Example 1:

Input: startTime = [1,2,3,3], endTime = [3,4,5,6], profit = [50,10,40,70]
Output: 120
Explanation: The subset chosen is the first and fourth job. 
Time range [1-3]+[3-6] , we get profit of 120 = 50 + 70.

Example 2:

Input: startTime = [1,2,3,4,6], endTime = [3,5,10,6,9], profit = [20,20,100,70,60]
Output: 150
Explanation: The subset chosen is the first, fourth and fifth job. 
Profit obtained 150 = 20 + 70 + 60.

Example 3:

Input: startTime = [1,1,1], endTime = [2,3,4], profit = [5,6,4]
Output: 6

 1 class Solution {
 2     public int jobScheduling(int[] startTime, int[] endTime, int[] profit) {
 3         int n = startTime.length;
 4         int[][] jobs = new int[n][3];
 5         for (int i = 0; i < n; i++) {
 6             jobs[i] = new int[] { startTime[i], endTime[i], profit[i] };
 7         }
 8         Arrays.sort(jobs, (a, b) -> a[1] - b[1]);
 9         TreeMap<Integer, Integer> dp = new TreeMap<>();
10         dp.put(0, 0);
11         for (int[] job : jobs) {
12             int cur = dp.floorEntry(job[0]).getValue() + job[2];
13             if (cur > dp.lastEntry().getValue())
14                 dp.put(job[1], cur);
15         }
16         return dp.lastEntry().getValue();
17     }
18 }

 

posted @ 2021-06-21 08:58  北叶青藤  阅读(73)  评论(0编辑  收藏  举报