Add solution 630

leetcode/0630.Course-Schedule-III/630. Course Schedule III.go

@@ -0,0 +1,39 @@
+package leetcode
+
+import (
+	"container/heap"
+	"sort"
+)
+
+func scheduleCourse(courses [][]int) int {
+	sort.Slice(courses, func(i, j int) bool {
+		return courses[i][1] < courses[j][1]
+	})
+	maxHeap, time := &Schedule{}, 0
+	heap.Init(maxHeap)
+	for _, c := range courses {
+		if time+c[0] <= c[1] {
+			time += c[0]
+			heap.Push(maxHeap, c[0])
+		} else if (*maxHeap).Len() > 0 && (*maxHeap)[0] > c[0] {
+			time -= heap.Pop(maxHeap).(int) - c[0]
+			heap.Push(maxHeap, c[0])
+		}
+	}
+	return (*maxHeap).Len()
+}
+
+type Schedule []int
+
+func (s Schedule) Len() int           { return len(s) }
+func (s Schedule) Less(i, j int) bool { return s[i] > s[j] }
+func (s Schedule) Swap(i, j int)      { s[i], s[j] = s[j], s[i] }
+func (s *Schedule) Pop() interface{} {
+	n := len(*s)
+	t := (*s)[n-1]
+	*s = (*s)[:n-1]
+	return t
+}
+func (s *Schedule) Push(x interface{}) {
+	*s = append(*s, x.(int))
+}

leetcode/0630.Course-Schedule-III/630. Course Schedule III_test.go

@@ -0,0 +1,42 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question630 struct {
+	para630
+	ans630
+}
+
+// para 是参数
+// one 代表第一个参数
+type para630 struct {
+	courses [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans630 struct {
+	one int
+}
+
+func Test_Problem630(t *testing.T) {
+
+	qs := []question630{
+
+		{
+			para630{[][]int{{100, 200}, {200, 1300}, {1000, 1250}, {2000, 3200}}},
+			ans630{3},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 630------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans630, q.para630
+		fmt.Printf("【input】:%v       【output】:%v\n", p, scheduleCourse(p.courses))
+	}
+	fmt.Printf("\n\n\n")
+}

leetcode/0630.Course-Schedule-III/README.md

@@ -0,0 +1,96 @@
+# [630. Course Schedule III](https://leetcode.com/problems/course-schedule-iii/)
+
+## 题目
+
+There are `n` different online courses numbered from `1` to `n`. You are given an array `courses` where `courses[i] = [durationi, lastDayi]` indicate that the `ith` course should be taken **continuously** for `durationi` days and must be finished before or on `lastDayi`.
+
+You will start on the `1st` day and you cannot take two or more courses simultaneously.
+
+Return *the maximum number of courses that you can take*.
+
+**Example 1:**
+
+```
+Input: courses = [[100,200],[200,1300],[1000,1250],[2000,3200]]
+Output: 3
+Explanation:
+There are totally 4 courses, but you can take 3 courses at most:
+First, take the 1st course, it costs 100 days so you will finish it on the 100th day, and ready to take the next course on the 101st day.
+Second, take the 3rd course, it costs 1000 days so you will finish it on the 1100th day, and ready to take the next course on the 1101st day.
+Third, take the 2nd course, it costs 200 days so you will finish it on the 1300th day.
+The 4th course cannot be taken now, since you will finish it on the 3300th day, which exceeds the closed date.
+
+```
+
+**Example 2:**
+
+```
+Input: courses = [[1,2]]
+Output: 1
+
+```
+
+**Example 3:**
+
+```
+Input: courses = [[3,2],[4,3]]
+Output: 0
+
+```
+
+**Constraints:**
+
+- `1 <= courses.length <= 104`
+- `1 <= durationi, lastDayi <= 104`
+
+## 题目大意
+
+这里有 n 门不同的在线课程，他们按从 1 到 n 编号。每一门课程有一定的持续上课时间（课程时间）t 以及关闭时间第 d 天。一门课要持续学习 t 天直到第 d 天时要完成，你将会从第 1 天开始。给出 n 个在线课程用 (t, d) 对表示。你的任务是找出最多可以修几门课。
+
+## 解题思路
+
+- 一般选课，任务的题目会涉及排序 + 贪心。此题同样如此。最多修几门课，采用贪心的思路。先将课程结束时间从小到大排序，优先选择结束时间靠前的课程，这样留给后面课程的时间越多，便可以修更多的课。对排好序的课程从前往后选课，不断累积时间。如果选择修当前课程，但是会超时，这时改调整了。对于已经选择的课程，都加入到最大堆中，遇到需要调整时，比较当前待考虑的课程时长是否比(堆中)已经选择课中时长最长的课时长短，即堆顶的课程时长短，剔除 pop 它，再选择这门时长短的课，并加入最大堆中。并更新累积时间。一层循环扫完所有课程，最终最大堆中包含课程的数目便是最多可以修的课程数。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"container/heap"
+	"sort"
+)
+
+func scheduleCourse(courses [][]int) int {
+	sort.Slice(courses, func(i, j int) bool {
+		return courses[i][1] < courses[j][1]
+	})
+	maxHeap, time := &Schedule{}, 0
+	heap.Init(maxHeap)
+	for _, c := range courses {
+		if time+c[0] <= c[1] {
+			time += c[0]
+			heap.Push(maxHeap, c[0])
+		} else if (*maxHeap).Len() > 0 && (*maxHeap)[0] > c[0] {
+			time -= heap.Pop(maxHeap).(int) - c[0]
+			heap.Push(maxHeap, c[0])
+		}
+	}
+	return (*maxHeap).Len()
+}
+
+type Schedule []int
+
+func (s Schedule) Len() int           { return len(s) }
+func (s Schedule) Less(i, j int) bool { return s[i] > s[j] }
+func (s Schedule) Swap(i, j int)      { s[i], s[j] = s[j], s[i] }
+func (s *Schedule) Pop() interface{} {
+	n := len(*s)
+	t := (*s)[n-1]
+	*s = (*s)[:n-1]
+	return t
+}
+func (s *Schedule) Push(x interface{}) {
+	*s = append(*s, x.(int))
+}
+```