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Commit fea22f8

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Merge pull request #389 from 0xff-dev/1834
Add solution and test-cases for problem 1834
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‎leetcode/1801-1900/1834.Single-Threaded-CPU/README.md‎

Lines changed: 34 additions & 14 deletions
Original file line numberDiff line numberDiff line change
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# [1834.Single-Threaded CPU][title]
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> [!WARNING|style:flat]
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> This question is temporarily unanswered if you have good ideas. Welcome to [Create Pull Request PR](https://github.com/kylesliu/awesome-golang-algorithm)
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## Description
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You are given `n` tasks labeled from `0` to `n - 1` represented by a 2D integer array `tasks`, where `tasks[i] = [enqueueTimei, processingTimei]` means that the i<sup>th</sup> task will be available to process at enqueueTime<sub>i</sub> and will take processingTime<sub>i</sub> to finish processing.
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You have a single-threaded CPU that can process **at most one** task at a time and will act in the following way:
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- If the CPU is idle and there are no available tasks to process, the CPU remains idle.
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- If the CPU is idle and there are available tasks, the CPU will choose the one with the **shortest processing time**. If multiple tasks have the same shortest processing time, it will choose the task with the smallest index.
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- Once a task is started, the CPU will **process the entire task** without stopping.
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- The CPU can finish a task then start a new one instantly.
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Return the order in which the CPU will process the tasks.
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**Example 1:**
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```
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Input: a = "11", b = "1"
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Output: "100"
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Input: tasks = [[1,2],[2,4],[3,2],[4,1]]
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Output: [0,2,3,1]
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Explanation: The events go as follows:
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- At time = 1, task 0 is available to process. Available tasks = {0}.
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- Also at time = 1, the idle CPU starts processing task 0. Available tasks = {}.
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- At time = 2, task 1 is available to process. Available tasks = {1}.
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- At time = 3, task 2 is available to process. Available tasks = {1, 2}.
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- Also at time = 3, the CPU finishes task 0 and starts processing task 2 as it is the shortest. Available tasks = {1}.
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- At time = 4, task 3 is available to process. Available tasks = {1, 3}.
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- At time = 5, the CPU finishes task 2 and starts processing task 3 as it is the shortest. Available tasks = {1}.
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- At time = 6, the CPU finishes task 3 and starts processing task 1. Available tasks = {}.
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- At time = 10, the CPU finishes task 1 and becomes idle.
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```
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## 题意
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> ...
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**Example 2:**
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## 题解
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### 思路1
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> ...
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Single-Threaded CPU
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```go
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```
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Input: tasks = [[7,10],[7,12],[7,5],[7,4],[7,2]]
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Output: [4,3,2,0,1]
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Explanation: The events go as follows:
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- At time = 7, all the tasks become available. Available tasks = {0,1,2,3,4}.
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- Also at time = 7, the idle CPU starts processing task 4. Available tasks = {0,1,2,3}.
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- At time = 9, the CPU finishes task 4 and starts processing task 3. Available tasks = {0,1,2}.
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- At time = 13, the CPU finishes task 3 and starts processing task 2. Available tasks = {0,1}.
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- At time = 18, the CPU finishes task 2 and starts processing task 0. Available tasks = {1}.
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- At time = 28, the CPU finishes task 0 and starts processing task 1. Available tasks = {}.
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- At time = 40, the CPU finishes task 1 and becomes idle.
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```
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## 结语
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package Solution
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func Solution(x bool) bool {
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import (
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"container/heap"
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"sort"
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)
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/*
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这道题开始是这样定义的结构体, 想用Pop后,更新costTime,然后在Fix,后面发现Fix的不太对
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type task1 struct {
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costTime
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tasks []task
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}
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*/
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type task struct {
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index int
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cost []int
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}
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type taskHeap []task
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func (t *taskHeap) Len() int {
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return len(*t)
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}
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func (t *taskHeap) Less(i, j int) bool {
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a := (*t)[i]
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b := (*t)[j]
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if a.cost[1] == b.cost[1] {
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return a.index < b.index
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}
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return a.cost[1] < b.cost[1]
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}
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func (t *taskHeap) Swap(i, j int) {
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(*t)[i], (*t)[j] = (*t)[j], (*t)[i]
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}
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func (t *taskHeap) Push(x interface{}) {
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*t = append(*t, x.(task))
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}
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func (t *taskHeap) Pop() interface{} {
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old := *t
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l := len(old)
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x := old[l-1]
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*t = old[:l-1]
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return x
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}
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func Solution(tasks [][]int) []int {
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tl := len(tasks)
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ans := make([]int, tl)
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taskList := make([]task, tl)
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for i := 0; i < len(tasks); i++ {
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taskList[i] = task{index: i, cost: tasks[i]}
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}
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h := taskHeap{}
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sort.Slice(taskList, func(i, j int) bool {
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return taskList[i].cost[0] < taskList[j].cost[0]
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})
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// wait first task
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costTime, done, i := 0, 0, 0
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for done < tl {
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if taskList[done].cost[0] > costTime {
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costTime = taskList[done].cost[0]
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}
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for ; done < tl && taskList[done].cost[0] <= costTime; done++ {
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heap.Push(&h, taskList[done])
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}
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for h.Len() > 0 && done < tl && costTime < taskList[done].cost[0] {
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x := heap.Pop(&h).(task)
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ans[i] = x.index
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i++
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costTime += x.cost[1]
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}
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}
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for h.Len() > 0 {
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x := heap.Pop(&h).(task)
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ans[i] = x.index
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i++
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}
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return ans
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}

‎leetcode/1801-1900/1834.Single-Threaded-CPU/Solution_test.go‎

Lines changed: 13 additions & 7 deletions
Original file line numberDiff line numberDiff line change
@@ -10,12 +10,18 @@ func TestSolution(t *testing.T) {
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// 测试用例
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cases := []struct {
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name string
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inputs bool
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expect bool
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inputs [][]int
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expect []int
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}{
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{"TestCase", true, true},
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{"TestCase", true, true},
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{"TestCase", false, false},
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{"TestCase1", [][]int{
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{1, 2}, {2, 4}, {3, 2}, {4, 1},
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}, []int{0, 2, 3, 1}},
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{"TestCase2", [][]int{
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{7, 10}, {7, 12}, {7, 5}, {7, 4}, {7, 2},
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}, []int{4, 3, 2, 0, 1}},
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{"TestCase3", [][]int{
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{19, 13}, {16, 9}, {21, 10}, {32, 25}, {37, 4}, {49, 24}, {2, 15}, {38, 41}, {37, 34}, {33, 6}, {45, 4}, {18, 18}, {46, 39}, {12, 24},
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}, []int{6, 1, 2, 9, 4, 10, 0, 11, 5, 13, 3, 8, 12, 7}},
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}
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// 开始测试
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}
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}
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//压力测试
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//压力测试
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func BenchmarkSolution(b *testing.B) {
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}
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//使用案列
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//使用案列
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func ExampleSolution() {
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}

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