★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★➤微信公众号：山青咏芝（shanqingyongzhi）➤博客园地址：山青咏芝（https://www.cnblogs.com/strengthen/）➤GitHub地址：https://github.com/strengthen/LeetCode➤原文地址：https://www.cnblogs.com/strengthen/p/10473321.html ➤如果链接不是山青咏芝的博客园地址，则可能是爬取作者的文章。➤原文已修改更新！强烈建议点击原文地址阅读！支持作者！支持原创！★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
There are n different online courses numbered from 1 to n. Each course has some duration(course length) t and closed on dth day. A course should be taken continuouslyfor t days and must be finished before or on the dth day. You will start at the 1st day.
Given n online courses represented by pairs (t,d), your task is to find the maximal number of courses that can be taken.
Example:
Input: [[100, 200], [200, 1300], [1000, 1250], [2000, 3200]]
Output: 3
Explanation:
There're 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.
Note:
The integer 1 <= d, t, n <= 10,000.You can't take two courses simultaneously.
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。
示例：
输入: [[100, 200], [200, 1300], [1000, 1250], [2000, 3200]]
输出: 3
解释:
这里一共有 4 门课程, 但是你最多可以修 3 门:
首先, 修第一门课时, 它要耗费 100 天，你会在第 100 天完成, 在第 101 天准备下门课。
第二, 修第三门课时, 它会耗费 1000 天，所以你将在第 1100 天的时候完成它, 以及在第 1101 天开始准备下门课程。
第三, 修第二门课时, 它会耗时 200 天，所以你将会在第 1300 天时完成它。
第四门课现在不能修，因为你将会在第 3300 天完成它，这已经超出了关闭日期。
提示:
整数 1 <= d, t, n <= 10,000 。你不能同时修两门课程。
Runtime: 1216 ms
Memory Usage: 19.6 MB
1 class Solution {
2 func scheduleCourse(_ courses: [[Int]]) ->
Int {
3 var courses =
courses
4 var curTime:Int =
0
5 var priorityQueue = PriorityQueue<Int>(order: >
)
6 courses.sort(by:{
7 (a:[Int],b:[Int]) -> Bool
in
8 return a[
1] < b[
1]
9 })
10 for course
in courses
11 {
12 curTime += course[
0]
13 priorityQueue.enqueue(course[
0])
14 if curTime > course[
1],let num:Int? =
priorityQueue.dequeue()
15 {
16 if num !=
nil
17 {
18 curTime -= num!
19 }
20 }
21 }
22 return priorityQueue.count
23 }
24 }
25
26 struct PriorityQueue<Element: Equatable>
{
27 private var heap: Heap<Element>
28
29 init(order: @escaping (Element, Element) ->
Bool) {
30 heap =
Heap(order: order)
31 }
32
33 var isEmpty: Bool {
34 return heap.isEmpty
35 }
36
37 var count:Int
38 {
39 return heap.count
40 }
41
42 var peek: Element?
{
43 return heap.peek
44 }
45
46 mutating func enqueue(_ element: Element) {
47 heap.insert(element)
48 }
49
50 mutating func dequeue() -> Element?
{
51 return heap.removePeek()
52 }
53 }
54
55 extension PriorityQueue: CustomStringConvertible {
56 var description: String {
57 return heap.description
58 }
59 }
60
61 struct Heap<Element: Equatable>
{
62 private(
set)
var elements: [Element] =
[]
63 private let order: (Element, Element) ->
Bool
64
65 init(order: @escaping (Element, Element) ->
Bool) {
66 self.order =
order
67 }
68
69 var isEmpty: Bool {
70 return elements.isEmpty
71 }
72
73 var count: Int {
74 return elements.count
75 }
76
77 var peek: Element?
{
78 return elements.first
79 }
80
81 func leftChildIndex(ofParentAt index: Int) ->
Int {
82 return 2 * index +
1
83 }
84
85 func rightChildIndex(ofParentAt index: Int) ->
Int {
86 return 2 * index +
2
87 }
88
89 func parentIndex(ofChildAt index: Int) ->
Int {
90 return (index -
1) /
2
91 }
92 }
93
94 extension Heap: CustomStringConvertible {
95 var description: String {
96 return elements.description
97 }
98 }
99
100 // MARK: - Remove & Insert
101 extension Heap {
102 @discardableResult
103 mutating func removePeek() -> Element?
{
104 guard !isEmpty
else {
105 return nil
106 }
107 elements.swapAt(
0, count -
1)
108 defer {
109 validateDown(
from:
0)
110 }
111 return elements.removeLast()
112 }
113
114 @discardableResult
115 mutating func remove(at index: Int) -> Element?
{
116 guard index < elements.count
else {
117 return nil
118 }
119 if index == elements.count -
1 {
120 return elements.removeLast()
121 }
else {
122 elements.swapAt(index, elements.count -
1)
123 defer {
124 validateDown(
from: index)
125 validateUp(
from: index)
126 }
127 return elements.removeLast()
128 }
129 }
130
131 mutating func insert(_ element: Element) {
132 elements.append(element)
133 validateUp(
from: elements.count -
1)
134 }
135
136 private mutating func validateUp(
from index: Int) {
137 var childIndex =
index
138 var parentIndex =
self.parentIndex(ofChildAt: childIndex)
139
140 while childIndex >
0 &&
141 order(elements[childIndex], elements[parentIndex]) {
142 elements.swapAt(childIndex, parentIndex)
143 childIndex =
parentIndex
144 parentIndex =
self.parentIndex(ofChildAt: childIndex)
145 }
146 }
147
148 private mutating func validateDown(
from index: Int) {
149 var parentIndex =
index
150 while true {
151 let leftIndex =
leftChildIndex(ofParentAt: parentIndex)
152 let rightIndex =
rightChildIndex(ofParentAt: parentIndex)
153 var targetParentIndex =
parentIndex
154
155 if leftIndex < count &&
156 order(elements[leftIndex], elements[targetParentIndex]) {
157 targetParentIndex =
leftIndex
158 }
159
160 if rightIndex < count &&
161 order(elements[rightIndex], elements[targetParentIndex]) {
162 targetParentIndex =
rightIndex
163 }
164
165 if targetParentIndex ==
parentIndex {
166 return
167 }
168
169 elements.swapAt(parentIndex, targetParentIndex)
170 parentIndex =
targetParentIndex
171 }
172 }
173 }
174
175 // MARK: - Search
176 extension Heap {
177 func index(of element: Element,
178 searchingFrom index: Int =
0) -> Int?
{
179 if index >=
count {
180 return nil
181 }
182 if order(element, elements[index]) {
183 return nil
184 }
185 if element ==
elements[index] {
186 return index
187 }
188
189 let leftIndex =
leftChildIndex(ofParentAt: index)
190 if let i =
self.index(of: element,
191 searchingFrom: leftIndex) {
192 return i
193 }
194
195 let rightIndex =
rightChildIndex(ofParentAt: index)
196 if let i =
self.index(of: element,
197 searchingFrom: rightIndex) {
198 return i
199 }
200
201 return nil
202 }
203 }
转载于:https://www.cnblogs.com/strengthen/p/10473321.html
