1235. Maximum Profit in Job Scheduling

# 1235. Maximum Profit in Job Scheduling#

## 题目 #

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 `startTime` , `endTime` and `profit` arrays, you need to output the maximum profit you can take such that there are no 2 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
``````

Constraints:

• `1 <= startTime.length == endTime.length == profit.length <= 5 * 10^4`
• `1 <= startTime[i] < endTime[i] <= 10^9`
• `1 <= profit[i] <= 10^4`

## 题目大意 #

• 1 <= startTime.length == endTime.length == profit.length <= 5 * 10^4
• 1 <= startTime[i] < endTime[i] <= 10^9
• 1 <= profit[i] <= 10^4

## 解题思路 #

• 给出一组任务，任务有开始时间，结束时间，和任务收益。一个任务开始还没有结束，中间就不能再安排其他任务。问如何安排任务，能使得最后收益最大？
• 一般任务的题目，区间的题目，都会考虑是否能排序。这一题可以先按照任务的结束时间从小到大排序，如果结束时间相同，则按照收益从小到大排序。`dp[i]` 代表前 `i` 份工作能获得的最大收益。初始值，`dp[0] = job[1].profit` 。对于任意一个任务 `i` ，看能否找到满足 `jobs[j].enTime <= jobs[j].startTime && j < i` 条件的 `j`，即查找 `upper_bound` 。由于 `jobs` 被我们排序了，所以这里可以使用二分搜索来查找。如果能找到满足条件的任务 j，那么状态转移方程是：`dp[i] = max(dp[i-1], jobs[i].profit)`。如果能找到满足条件的任务 j，那么状态转移方程是：`dp[i] = max(dp[i-1], dp[low]+jobs[i].profit)`。最终求得的解在 `dp[len(startTime)-1]` 中。

## 代码 #

``````
package leetcode

import "sort"

type job struct {
startTime int
endTime   int
profit    int
}

func jobScheduling(startTime []int, endTime []int, profit []int) int {
jobs, dp := []job{}, make([]int, len(startTime))
for i := 0; i < len(startTime); i++ {
jobs = append(jobs, job{startTime: startTime[i], endTime: endTime[i], profit: profit[i]})
}
sort.Sort(sortJobs(jobs))
dp[0] = jobs[0].profit
for i := 1; i < len(jobs); i++ {
low, high := 0, i-1
for low < high {
mid := low + (high-low)>>1
if jobs[mid+1].endTime <= jobs[i].startTime {
low = mid + 1
} else {
high = mid
}
}
if jobs[low].endTime <= jobs[i].startTime {
dp[i] = max(dp[i-1], dp[low]+jobs[i].profit)
} else {
dp[i] = max(dp[i-1], jobs[i].profit)
}
}
return dp[len(startTime)-1]
}

type sortJobs []job

func (s sortJobs) Len() int {
return len(s)
}
func (s sortJobs) Less(i, j int) bool {
if s[i].endTime == s[j].endTime {
return s[i].profit < s[j].profit
}
return s[i].endTime < s[j].endTime
}
func (s sortJobs) Swap(i, j int) {
s[i], s[j] = s[j], s[i]
}

``````

Apr 8, 2023