汇总：模拟卷Leetcode 题解汇总

539. 最小时间差

给定一个 24 小时制（小时:分钟 “HH:MM”）的时间列表，找出列表中任意两个时间的最小时间差并以分钟数表示。

示例 1：

输入：timePoints = [“23:59”,“00:00”]
输出：1
示例 2：

输入：timePoints = [“00:00”,“23:59”,“00:00”]
输出：0

提示：

2 <= timePoints <= 2 * 104
timePoints[i] 格式为 “HH:MM”

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/minimum-time-difference
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

代码：

from leetcode_python.utils import *


class Solution:
    def diff(self,time1,time2):
        """return time2-time1"""
        # h1,m1 = int(time1[0:2]),int(time1[3:5])
        # h2,m2 = int(time2[0:2]),int(time2[3:5])
        h1,m1 = map(int,time1.split(':'))
        h2,m2 = map(int,time2.split(':'))
        return (h2-h1)*60+m2-m1


    def findMinDifference(self, timePoints: List[str]) -> int:
        timePoints.sort()
        res = self.diff(timePoints[-1],str(int(timePoints[0][:2])+24)+timePoints[0][2:])
        for id,t in enumerate(timePoints):
            if 0==id:continue
            res = min(res,self.diff(timePoints[id-1],t))
        return res


def test(data_test):
    s = Solution()
    data = data_test  # normal
    # data = [list2node(data_test[0])]  # list转node
    return s.findMinDifference(*data)


def test_obj(data_test):
    result = [None]
    obj = Solution(*data_test[1][0])
    for fun, data in zip(data_test[0][1::], data_test[1][1::]):
        if data:
            res = obj.__getattribute__(fun)(*data)
        else:
            res = obj.__getattribute__(fun)()
        result.append(res)
    return result


if __name__ == '__main__':
    datas = [
        [["00:00","23:59","00:00"]],
    ]
    for data_test in datas:
        t0 = time.time()
        print('-' * 50)
        print('input:', data_test)
        print('output:', test(data_test))
        print(f'use time:{time.time() - t0}s')

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备注：
GitHub：https://github.com/monijuan/leetcode_python

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leetcode_python.utils详见汇总页说明
先刷的题，之后用脚本生成的blog，如果有错请留言，我看到了会修改的！谢谢！