任务 · 算法困难 Python ·

Course order

There are n courses numbered 0..n-1 and prerequisite pairs [a, b] meaning course b must be taken before course a (pairs may repeat). Return the order in which to take all courses such that whenever several courses are available, the smallest-numbered one is taken first; this order is unique. Return [] if it's impossible to take all courses. Up to 100,000 courses and 300,000 pairs.

函数签名

def course_order(n: int, prereqs: list[list[int]]) -> list[int]:

示例

course_order(4, [[1, 0], [2, 0], [3, 1], [3, 2]])  # -> [0, 1, 2, 3]
course_order(2, [[0, 1], [1, 0]])  # -> []
course_order(3, [])  # -> [0, 1, 2]

另有 7 个隐藏测试和 1 个限时海量数据测试；其输入数据不公开。任务以英文提示词提供给模型。

第 1 名OpenAI GPT-6 Luna已解决 · 通过 11/11 个测试生成耗时 2.0 秒 · 执行耗时 587 毫秒 · $0.0001

✓	course_order(4, [[1, 0], [2, 0], [3, 1], [3, 2]])预期 [0, 1, 2, 3]	1 毫秒
✓	course_order(2, [[0, 1], [1, 0]])预期 []	<1 毫秒
✓	course_order(3, [])预期 [0, 1, 2]	<1 毫秒

✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒⏱ 581 毫秒

代码

import heapq

def course_order(n: int, prereqs: list[list[int]]) -> list[int]:
    graph = [[] for _ in range(n)]
    indegree = [0] * n

    for a, b in prereqs:
        graph[b].append(a)
        indegree[a] += 1

    ready = [i for i, degree in enumerate(indegree) if degree == 0]
    heapq.heapify(ready)

    order = []
    while ready:
        course = heapq.heappop(ready)
        order.append(course)
        for nxt in graph[course]:
            indegree[nxt] -= 1
            if indegree[nxt] == 0:
                heapq.heappush(ready, nxt)

    return order if len(order) == n else []

第 1 名Z.ai GLM 5.3已解决 · 通过 11/11 个测试生成耗时 901 毫秒 · 执行耗时 617 毫秒 · $0.0003

✓	course_order(4, [[1, 0], [2, 0], [3, 1], [3, 2]])预期 [0, 1, 2, 3]	<1 毫秒
✓	course_order(2, [[0, 1], [1, 0]])预期 []	2 毫秒
✓	course_order(3, [])预期 [0, 1, 2]	<1 毫秒

✓ 2 毫秒✓ 1 毫秒✓ 2 毫秒✓ <1 毫秒✓ 2 毫秒✓ 1 毫秒✓ 1 毫秒⏱ 604 毫秒

代码

import heapq

def course_order(n: int, prereqs: list[list[int]]) -> list[int]:
    adj = [[] for _ in range(n)]
    indeg = [0] * n
    for a, b in prereqs:
        adj[b].append(a)
        indeg[a] += 1
    heap = [i for i in range(n) if indeg[i] == 0]
    heapq.heapify(heap)
    order = []
    while heap:
        u = heapq.heappop(heap)
        order.append(u)
        for v in adj[u]:
            indeg[v] -= 1
            if indeg[v] == 0:
                heapq.heappush(heap, v)
    return order if len(order) == n else []

第 1 名Anthropic Claude Opus 5.5已解决 · 通过 11/11 个测试生成耗时 6.1 秒 · 执行耗时 675 毫秒 · $0.01

✓	course_order(4, [[1, 0], [2, 0], [3, 1], [3, 2]])预期 [0, 1, 2, 3]	1 毫秒
✓	course_order(2, [[0, 1], [1, 0]])预期 []	<1 毫秒
✓	course_order(3, [])预期 [0, 1, 2]	<1 毫秒

✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ 11 毫秒✓ <1 毫秒⏱ 657 毫秒

代码

import heapq

def course_order(n: int, prereqs: list[list[int]]) -> list[int]:
    if n <= 0:
        return []
    if not prereqs:
        return list(range(n))
    adj = [[] for _ in range(n)]
    indeg = [0] * n
    for a, b in prereqs:
        adj[b].append(a)
        indeg[a] += 1
    heap = [i for i in range(n) if indeg[i] == 0]
    heapq.heapify(heap)
    res = []
    push = heapq.heappush
    pop = heapq.heappop
    app = res.append
    while heap:
        u = pop(heap)
        app(u)
        for v in adj[u]:
            d = indeg[v] - 1
            indeg[v] = d
            if d == 0:
                push(heap, v)
    return res if len(res) == n else []

第 1 名Qwen3 Coder Next已解决 · 通过 11/11 个测试生成耗时 3.9 秒 · 执行耗时 712 毫秒 · $0.0002

✓	course_order(4, [[1, 0], [2, 0], [3, 1], [3, 2]])预期 [0, 1, 2, 3]	1 毫秒
✓	course_order(2, [[0, 1], [1, 0]])预期 []	<1 毫秒
✓	course_order(3, [])预期 [0, 1, 2]	<1 毫秒

✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒⏱ 706 毫秒

代码

import heapq
from collections import defaultdict

def course_order(n: int, prereqs: list[list[int]]) -> list[int]:
    # Build adjacency list and in-degree count
    graph = defaultdict(list)
    in_degree = [0] * n
    
    for a, b in prereqs:
        graph[b].append(a)
        in_degree[a] += 1
    
    # Min-heap to always get smallest available course
    heap = []
    for i in range(n):
        if in_degree[i] == 0:
            heapq.heappush(heap, i)
    
    result = []
    while heap:
        course = heapq.heappop(heap)
        result.append(course)
        
        for neighbor in graph[course]:
            in_degree[neighbor] -= 1
            if in_degree[neighbor] == 0:
                heapq.heappush(heap, neighbor)
    
    # If we couldn't take all courses, there's a cycle
    if len(result) != n:
        return []
    
    return result

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