[LeetCode] 2446. Determine if Two Events Have Conflict

You are given two arrays of strings that represent two inclusive events that happened on the same day, event1 and event2, where:

event1 = [startTime1, endTime1] and
event2 = [startTime2, endTime2].

Event times are valid 24 hours format in the form of HH:MM.

A conflict happens when two events have some non-empty intersection (i.e., some moment is common to both events).

Return true if there is a conflict between two events. Otherwise, return false.

Example 1:

Input: event1 = ["01:15","02:00"], event2 = ["02:00","03:00"]
Output: true
Explanation: The two events intersect at time 2:00.

Example 2:

Input: event1 = ["01:00","02:00"], event2 = ["01:20","03:00"]
Output: true
Explanation: The two events intersect starting from 01:20 to 02:00.

Example 3:

Input: event1 = ["10:00","11:00"], event2 = ["14:00","15:00"]
Output: false
Explanation: The two events do not intersect.

Constraints:

evnet1.length == event2.length == 2.
event1[i].length == event2[i].length == 5
startTime1 <= endTime1
startTime2 <= endTime2
All the event times follow the HH:MM format.

判断两个事件是否存在冲突。

给你两个字符串数组 event1 和 event2 ，表示发生在同一天的两个闭区间时间段事件，其中：

event1 = [startTime1, endTime1] 且
event2 = [startTime2, endTime2]
事件的时间为有效的 24 小时制且按 HH:MM 格式给出。

当两个事件存在某个非空的交集时（即，某些时刻是两个事件都包含的），则认为出现冲突。

如果两个事件之间存在冲突，返回 true ；否则，返回 false 。

来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/determine-if-two-events-have-conflict
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这道题不涉及算法，但是这道题的做法很巧妙。如果两个 event 的时间冲突，那么一定满足 startTime1 <= endTime2 && endTime1 <= startTime2，画两个平行的线段就知道了，如下图。为什么要考虑 startTime1 <= endTime2 呢，因为有可能 event2 的时间是小于 event1 的。

[startTime1, endTime1]

[startTime2, endTime2]

时间O(1)

空间O(1)

Java实现

1 class Solution {
2     public boolean haveConflict(String[] event1, String[] event2) {
3         return event1[0].compareTo(event2[1]) <= 0 && event2[0].compareTo(event1[1]) <= 0;
4     }
5 }