diff --git a/Algorithms/1235. Maximum Profit in Job Scheduling/1235. Maximum Profit in Job Scheduling.go b/Algorithms/1235. Maximum Profit in Job Scheduling/1235. Maximum Profit in Job Scheduling.go new file mode 100644 index 00000000..91eb4900 --- /dev/null +++ b/Algorithms/1235. Maximum Profit in Job Scheduling/1235. Maximum Profit in Job Scheduling.go @@ -0,0 +1,50 @@ +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] +} diff --git a/Algorithms/1235. Maximum Profit in Job Scheduling/1235. Maximum Profit in Job Scheduling_test.go b/Algorithms/1235. Maximum Profit in Job Scheduling/1235. Maximum Profit in Job Scheduling_test.go new file mode 100644 index 00000000..65b49a7d --- /dev/null +++ b/Algorithms/1235. Maximum Profit in Job Scheduling/1235. Maximum Profit in Job Scheduling_test.go @@ -0,0 +1,54 @@ +package leetcode + +import ( + "fmt" + "testing" +) + +type question1235 struct { + para1235 + ans1235 +} + +// para 是参数 +// one 代表第一个参数 +type para1235 struct { + startTime []int + endTime []int + profit []int +} + +// ans 是答案 +// one 代表第一个答案 +type ans1235 struct { + one int +} + +func Test_Problem1235(t *testing.T) { + + qs := []question1235{ + + question1235{ + para1235{[]int{1, 2, 3, 3}, []int{3, 4, 5, 6}, []int{50, 10, 40, 70}}, + ans1235{120}, + }, + + question1235{ + para1235{[]int{1, 2, 3, 4, 6}, []int{3, 5, 10, 6, 9}, []int{20, 20, 100, 70, 60}}, + ans1235{150}, + }, + + question1235{ + para1235{[]int{1, 1, 1}, []int{2, 3, 4}, []int{5, 6, 4}}, + ans1235{6}, + }, + } + + fmt.Printf("------------------------Leetcode Problem 1235------------------------\n") + + for _, q := range qs { + _, p := q.ans1235, q.para1235 + fmt.Printf("【input】:%v 【output】:%v\n", p, jobScheduling(p.startTime, p.endTime, p.profit)) + } + fmt.Printf("\n\n\n") +} diff --git a/Algorithms/1235. Maximum Profit in Job Scheduling/README.md b/Algorithms/1235. Maximum Profit in Job Scheduling/README.md new file mode 100755 index 00000000..546c649e --- /dev/null +++ b/Algorithms/1235. Maximum Profit in Job Scheduling/README.md @@ -0,0 +1,60 @@ +# [1235. Maximum Profit in Job Scheduling](https://leetcode.com/problems/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:** + +![https://assets.leetcode.com/uploads/2019/10/10/sample1_1584.png](https://assets.leetcode.com/uploads/2019/10/10/sample1_1584.png) + + 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:** + +![https://assets.leetcode.com/uploads/2019/10/10/sample22_1584.png](https://assets.leetcode.com/uploads/2019/10/10/sample22_1584.png) + + 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:** + +![https://assets.leetcode.com/uploads/2019/10/10/sample3_1584.png](https://assets.leetcode.com/uploads/2019/10/10/sample3_1584.png) + + 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 + + +你打算利用空闲时间来做兼职工作赚些零花钱。这里有 n 份兼职工作，每份工作预计从 startTime[i] 开始到 endTime[i] 结束，报酬为 profit[i]。给你一份兼职工作表，包含开始时间 startTime，结束时间 endTime 和预计报酬 profit 三个数组，请你计算并返回可以获得的最大报酬。注意，时间上出现重叠的 2 份工作不能同时进行。如果你选择的工作在时间 X 结束，那么你可以立刻进行在时间 X 开始的下一份工作。 + + +提示： + +- 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。