1834.Single-Threaded CPU
===

###### tags: `Medium`,`Array`,`Heap`,`Sorting`

[1834. Single-Threaded CPU](https://leetcode.com/problems/single-threaded-cpu/)

### 題目描述

You are given `n` tasks labeled from `0` to `n - 1` represented by a 2D integer array tasks, where `tasks[i]` = [$enqueueTime_i$, $processingTime_i$] means that the i^th^ task will be available to process at $enqueueTime_i$ and will take $processingTime_i$ to finish processing.

You have a single-threaded CPU that can process at most one task at a time and will act in the following way:

* If the CPU is idle and there are no available tasks to process, the CPU remains idle.
* 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.
* Once a task is started, the CPU will **process the entire task** without stopping.
* The CPU can finish a task then start a new one instantly.

Return *the order in which the CPU will process the tasks.*

### 範例

**Example 1:**

Input: tasks = [[1,2],[2,4],[3,2],[4,1]]
Output: [0,2,3,1]
Explanation: The events go as follows: 
- At time = 1, task 0 is available to process. Available tasks = {0}.
- Also at time = 1, the idle CPU starts processing task 0. Available tasks = {}.
- At time = 2, task 1 is available to process. Available tasks = {1}.
- At time = 3, task 2 is available to process. Available tasks = {1, 2}.
- Also at time = 3, the CPU finishes task 0 and starts processing task 2 as it is the shortest. Available tasks = {1}.
- At time = 4, task 3 is available to process. Available tasks = {1, 3}.
- At time = 5, the CPU finishes task 2 and starts processing task 3 as it is the shortest. Available tasks = {1}.
- At time = 6, the CPU finishes task 3 and starts processing task 1. Available tasks = {}.
- At time = 10, the CPU finishes task 1 and becomes idle.

**Example 2:**

Input: tasks = [[7,10],[7,12],[7,5],[7,4],[7,2]]
Output: [4,3,2,0,1]
Explanation: The events go as follows:
- At time = 7, all the tasks become available. Available tasks = {0,1,2,3,4}.
- Also at time = 7, the idle CPU starts processing task 4. Available tasks = {0,1,2,3}.
- At time = 9, the CPU finishes task 4 and starts processing task 3. Available tasks = {0,1,2}.
- At time = 13, the CPU finishes task 3 and starts processing task 2. Available tasks = {0,1}.
- At time = 18, the CPU finishes task 2 and starts processing task 0. Available tasks = {1}.
- At time = 28, the CPU finishes task 0 and starts processing task 1. Available tasks = {}.
- At time = 40, the CPU finishes task 1 and becomes idle.

**Constraints**:

* `tasks.length` == `n`
* 1 <= `n` <= 10^5^
* 1 <= $enqueueTime_i$, $processingTime_i$ <= 10^9^

### 解答

#### Python

class Solution:
    def getOrder(self, tasks: List[List[int]]) -> List[int]:
        from queue import PriorityQueue
        task_cnt, current_time = 0, 0
        tasks = sorted([x+[i]
                       for i, x in enumerate(tasks)], key=lambda x: x[0])

        # print(tasks)

        current_time = tasks[0][0]
        pq = PriorityQueue()
        ans = []
        enq_idx = 0
        while task_cnt < len(tasks):
            while enq_idx < len(tasks) and tasks[enq_idx][0] <= current_time:
                pq.put((tasks[enq_idx][1]*100000 +
                       tasks[enq_idx][2], tasks[enq_idx]))
                # If multiple tasks have the same shortest processing time, it will choose the task with the smallest index.
                enq_idx += 1
                # print('#',flush=True)

            if not pq.empty():
                Q = pq.get()  # with lowest process time
            else:
                current_time = tasks[enq_idx][0]
                continue

            current_time += Q[1][1]

            ans.append(Q[1][2])
            # print(ans[-1],Q, current_time, ans)
            task_cnt += 1
            # print('@',flush=True)

        return ans

> [name=玉山][time=Thu, Dec 29, 2022 12:16 PM]
### Reference

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