添加 problem 1235

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]
+}

Algorithms/1235. Maximum Profit in Job Scheduling/1235. Maximum Profit in Job Scheduling_test.go

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+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")
+}

Algorithms/1235. Maximum Profit in Job Scheduling/README.md

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+# [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。