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113 lines
3.7 KiB
Markdown
113 lines
3.7 KiB
Markdown
# [1353. Maximum Number of Events That Can Be Attended](https://leetcode.com/problems/maximum-number-of-events-that-can-be-attended/)
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## 题目
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Given an array of `events` where `events[i] = [startDayi, endDayi]`. Every event `i` starts at `startDayi` and ends at `endDayi`.
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You can attend an event `i` at any day `d` where `startTimei <= d <= endTimei`. Notice that you can only attend one event at any time `d`.
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Return *the maximum number of events* you can attend.
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**Example 1:**
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```
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Input: events = [[1,2],[2,3],[3,4]]
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Output: 3
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Explanation: You can attend all the three events.
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One way to attend them all is as shown.
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Attend the first event on day 1.
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Attend the second event on day 2.
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Attend the third event on day 3.
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```
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**Example 2:**
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```
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Input: events= [[1,2],[2,3],[3,4],[1,2]]
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Output: 4
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```
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**Example 3:**
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```
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Input: events = [[1,4],[4,4],[2,2],[3,4],[1,1]]
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Output: 4
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```
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**Example 4:**
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```
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Input: events = [[1,100000]]
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Output: 1
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```
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**Example 5:**
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```
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Input: events = [[1,1],[1,2],[1,3],[1,4],[1,5],[1,6],[1,7]]
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Output: 7
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```
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**Constraints:**
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- `1 <= events.length <= 10^5`
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- `events[i].length == 2`
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- `1 <= startDayi <= endDayi <= 10^5`
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## 题目大意
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给你一个数组 events，其中 events[i] = [startDayi, endDayi] ，表示会议 i 开始于 startDayi ，结束于 endDayi 。你可以在满足 startDayi <= d <= endDayi 中的任意一天 d 参加会议 i 。注意，一天只能参加一个会议。请你返回你可以参加的 最大 会议数目。
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## 解题思路
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- 关于会议安排，活动安排这类题，第一直觉是贪心问题。先按照会议开始时间从小到大排序，如果开始时间相同，再按照结束时间从小到大排序。贪心策略是，优先选择参加早结束的会议。因为一个结束时间晚的会议，代表这个会议持续时间长，先参加马上要结束的会议，这样可以参加更多的会议。
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- 注意题目给的数据代表的是天数。比较大小的时候最好转换成坐标轴上的坐标点。例如 [1,2] 代表这个会议持续 2 天，如果在坐标轴上表示，是 [0,2]，0-1 表示第一天，1-2 表示第二天。所以比较会议时需要把开始时间减一。选定了这个会议以后记得要把这一天排除，例如选择了第二天，那么下次对比起始时间需要从坐标 2 开始，因为第二天的时间范围是 1-2，所以下一轮比较会议前需要把开始时间加一。从左往右依次扫描各个会议时间段，选择结束时间大于起始时间的会议，不断累加次数，扫描完所有会议，最终结果即为可参加的最大会议数。
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- 测试数据中有一组很恶心的数据，见 test 文件中最后一组数据。这组数据在同一天叠加了多个会议，并且起始时间完全一致。这种特殊情况需要加判断条件排除，见下面代码 continue 条件。
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## 代码
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```go
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package leetcode
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import (
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"sort"
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)
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func maxEvents(events [][]int) int {
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sort.Slice(events, func(i, j int) bool {
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if events[i][0] == events[j][0] {
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return events[i][1] < events[j][1]
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}
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return events[i][0] < events[j][0]
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})
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attended, current := 1, events[0]
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for i := 1; i < len(events); i++ {
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prev, event := events[i-1], events[i]
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if event[0] == prev[0] && event[1] == prev[1] && event[1] == event[0] {
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continue
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}
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start, end := max(current[0], event[0]-1), max(current[1], event[1])
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if end-start > 0 {
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current[0] = start + 1
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current[1] = end
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attended++
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}
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}
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return attended
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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``` |