如何grepl搜索字符串中的最大和最小计时？

import re

def find_max_min_time(string):
    # 使用正则表达式匹配"HH:MM:SS"格式的计时
    pattern = r'\d{2}:\d{2}:\d{2}'
    times = re.findall(pattern, string)
    
    if times:
        # 将时间转换成秒数
        seconds = [int(time.split(':')[0])*3600 + int(time.split(':')[1])*60 + int(time.split(':')[2]) for time in times]
        max_time = max(seconds)
        min_time = min(seconds)
        
        # 将秒数转换回"HH:MM:SS"格式
        max_time_str = f"{max_time//3600:02d}:{max_time//60%60:02d}:{max_time%60:02d}"
        min_time_str = f"{min_time//3600:02d}:{min_time//60%60:02d}:{min_time%60:02d}"
        
        return f"最大计时：{max_time_str}，最小计时：{min_time_str}"
    else:
        return "未找到匹配的计时"

# 测试示例
string = "这是一个示例字符串，其中包含计时01:23:45和02:34:56"
result = find_max_min_time(string)
print(result)

最大计时：02:34:56，最小计时：01:23:45