任务 · 算法中等 Python ·

Merge intervals

Each interval [a, b] with a ≤ b is closed. Merge all intervals that overlap or touch ([1, 3] and [3, 5] merge into [1, 5]; [1, 2] and [3, 4] stay apart). Return the merged intervals sorted by start. The input is in no particular order and can hold up to 200,000 intervals.

函数签名

def merge_intervals(intervals: list[list[int]]) -> list[list[int]]:

示例

merge_intervals([[1, 3], [2, 6], [8, 10], [15, 18]])  # -> [[1, 6], [8, 10], [15, 18]]
merge_intervals([[1, 4], [4, 5]])  # -> [[1, 5]]

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-6.1 Terra已解决 · 通过 10/10 个测试生成耗时 2.2 秒 · 执行耗时 582 毫秒 · $0.0009

✓	merge_intervals([[1, 3], [2, 6], [8, 10], [15, 18]])预期 [[1, 6], [8, 10], [15, 18]]	<1 毫秒
✓	merge_intervals([[1, 4], [4, 5]])预期 [[1, 5]]	<1 毫秒

✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒⏱ 576 毫秒

代码

def merge_intervals(intervals: list[list[int]]) -> list[list[int]]:
    if not intervals:
        return []

    ordered = sorted(intervals, key=lambda interval: interval[0])
    start, end = ordered[0]
    merged = []

    for a, b in ordered[1:]:
        if a <= end:
            if b > end:
                end = b
        else:
            merged.append([start, end])
            start, end = a, b

    merged.append([start, end])
    return merged

第 1 名SpaceXAI Grok Build 0.1已解决 · 通过 10/10 个测试生成耗时 5.0 秒 · 执行耗时 643 毫秒 · $0.0024

✓	merge_intervals([[1, 3], [2, 6], [8, 10], [15, 18]])预期 [[1, 6], [8, 10], [15, 18]]	2 毫秒
✓	merge_intervals([[1, 4], [4, 5]])预期 [[1, 5]]	1 毫秒

✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒⏱ 635 毫秒

代码

def merge_intervals(intervals: list[list[int]]) -> list[list[int]]:
    if not intervals:
        return []
    sorted_intervals = sorted(intervals, key=lambda x: x[0])
    merged = [sorted_intervals[0][:]]
    for current in sorted_intervals[1:]:
        prev = merged[-1]
        if current[0] <= prev[1]:
            prev[1] = max(prev[1], current[1])
        else:
            merged.append(current[:])
    return merged

第 1 名Google Gemini 3.8 Flash已解决 · 通过 10/10 个测试生成耗时 3.4 秒 · 执行耗时 673 毫秒 · $0.0007

✓	merge_intervals([[1, 3], [2, 6], [8, 10], [15, 18]])预期 [[1, 6], [8, 10], [15, 18]]	<1 毫秒
✓	merge_intervals([[1, 4], [4, 5]])预期 [[1, 5]]	<1 毫秒

✓ <1 毫秒✓ 1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒⏱ 666 毫秒

代码

def merge_intervals(intervals: list[list[int]]) -> list[list[int]]:
    if not intervals:
        return []
    
    intervals.sort(key=lambda x: x[0])
    
    merged = [intervals[0]]
    for start, end in intervals[1:]:
        last_end = merged[-1][1]
        if start <= last_end:
            if end > last_end:
                merged[-1][1] = end
        else:
            merged.append([start, end])
            
    return merged

第 1 名Qwen3 Coder Next已解决 · 通过 10/10 个测试生成耗时 2.6 秒 · 执行耗时 687 毫秒 · $0.0002

✓	merge_intervals([[1, 3], [2, 6], [8, 10], [15, 18]])预期 [[1, 6], [8, 10], [15, 18]]	1 毫秒
✓	merge_intervals([[1, 4], [4, 5]])预期 [[1, 5]]	1 毫秒

✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ 1 毫秒✓ 1 毫秒✓ 2 毫秒✓ 2 毫秒⏱ 676 毫秒

代码

def merge_intervals(intervals: list[list[int]]) -> list[list[int]]:
    if not intervals:
        return []
    
    # Sort intervals by their start values
    intervals.sort(key=lambda x: x[0])
    
    merged = []
    current_start, current_end = intervals[0]
    
    for i in range(1, len(intervals)):
        start, end = intervals[i]
        # If the current interval overlaps or touches the previous one
        if start <= current_end:
            # Merge them by updating the end if needed
            current_end = max(current_end, end)
        else:
            # No overlap, add the previous merged interval to result
            merged.append([current_start, current_end])
            current_start, current_end = start, end
    
    # Add the last interval
    merged.append([current_start, current_end])
    
    return merged

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