如何在python中查找列表中的第二大数

def find_second_largest(nums):
    sorted_nums = sorted(nums)
    return sorted_nums[-2] if len(sorted_nums) >= 2 else None

# 调用示例
numbers = [10, 5, 8, 12, 3]
second_largest = find_second_largest(numbers)
print(second_largest)  # 输出 10

def find_second_largest(nums):
    largest = float('-inf')
    second_largest = float('-inf')
    for num in nums:
        if num > largest:
            second_largest = largest
            largest = num
        elif num > second_largest and num != largest:
            second_largest = num
    return second_largest if second_largest != float('-inf') else None

# 调用示例
numbers = [10, 5, 8, 12, 3]
second_largest = find_second_largest(numbers)
print(second_largest)  # 输出 10

def find_second_largest(nums):
    unique_nums = set(nums)
    if len(unique_nums) < 2:
        return None
    unique_nums.remove(max(unique_nums))
    return max(unique_nums)

# 调用示例
numbers = [10, 5, 8, 12, 3]
second_largest = find_second_largest(numbers)
print(second_largest)  # 输出 10