要求
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"
核心代码
class Solution:
def findMinDifference(self, timePoints: List[str]) -> int:
n = len(timePoints)
res = []
for i in range(n):
res.append(int(timePoints[i][0:2])* 60 + int(timePoints[i][3:5]))
res.sort()
tmp = res[0] + 24 * 60 - res[n - 1]
for i in range(1,n):
tmp = min(tmp,res[i] - res[i - 1])
return tmp
解题思路:我们先将所有的时间字符串转换成整数分钟时间,然后我们对列表进行排序,排完序后前后相减保留最小值即可,比较简单。