将列表中具有特定容差的连续整数进行分组Python3

def group_consecutive_integers(lst, tolerance):
    if not lst:
        return []

    groups = []
    current_group = [lst[0]]

    for i in range(1, len(lst)):
        if abs(lst[i] - lst[i - 1]) <= tolerance:
            current_group.append(lst[i])
        else:
            groups.append(current_group)
            current_group = [lst[i]]

    groups.append(current_group)
    return groups

# 示例用法
lst = [1, 2, 3, 6, 7, 10, 11, 12]
tolerance = 2
result = group_consecutive_integers(lst, tolerance)
print(result)  # 输出: [[1, 2, 3], [6, 7], [10, 11, 12]]