Add solution 1353

leetcode/1353.Maximum-Number-of-Events-That-Can-Be-Attended/1353. Maximum Number of Events That Can Be Attended.go

@@ -0,0 +1,35 @@
+package leetcode
+
+import (
+	"sort"
+)
+
+func maxEvents(events [][]int) int {
+	sort.Slice(events, func(i, j int) bool {
+		if events[i][0] == events[j][0] {
+			return events[i][1] < events[j][1]
+		}
+		return events[i][0] < events[j][0]
+	})
+	attended, current := 1, events[0]
+	for i := 1; i < len(events); i++ {
+		prev, event := events[i-1], events[i]
+		if event[0] == prev[0] && event[1] == prev[1] && event[1] == event[0] {
+			continue
+		}
+		start, end := max(current[0], event[0]-1), max(current[1], event[1])
+		if end-start > 0 {
+			current[0] = start + 1
+			current[1] = end
+			attended++
+		}
+	}
+	return attended
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}

leetcode/1353.Maximum-Number-of-Events-That-Can-Be-Attended/1353. Maximum Number of Events That Can Be Attended_test.go

@@ -0,0 +1,67 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1353 struct {
+	para1353
+	ans1353
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1353 struct {
+	events [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1353 struct {
+	one int
+}
+
+func Test_Problem1353(t *testing.T) {
+
+	qs := []question1353{
+
+		{
+			para1353{[][]int{{1, 2}, {2, 3}, {3, 4}}},
+			ans1353{3},
+		},
+
+		{
+			para1353{[][]int{{1, 4}, {4, 4}, {2, 2}, {3, 4}, {1, 1}}},
+			ans1353{4},
+		},
+
+		{
+			para1353{[][]int{{1, 100000}}},
+			ans1353{1},
+		},
+
+		{
+			para1353{[][]int{{1, 1}, {2, 2}, {1, 3}, {1, 4}, {1, 5}, {1, 6}, {1, 7}}},
+			ans1353{7},
+		},
+
+		{
+			para1353{[][]int{{1, 2}, {2, 2}, {3, 3}, {3, 4}, {3, 4}}},
+			ans1353{4},
+		},
+
+		{
+			para1353{[][]int{{1, 10}, {2, 2}, {2, 2}, {2, 2}, {2, 2}}},
+			ans1353{2},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1353------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1353, q.para1353
+		fmt.Printf("【input】:%v       【output】:%v\n", p, maxEvents(p.events))
+	}
+	fmt.Printf("\n\n\n")
+}

leetcode/1353.Maximum-Number-of-Events-That-Can-Be-Attended/README.md

@@ -0,0 +1,113 @@
+# [1353. Maximum Number of Events That Can Be Attended](https://leetcode.com/problems/maximum-number-of-events-that-can-be-attended/)
+
+
+## 题目
+
+Given an array of `events` where `events[i] = [startDayi, endDayi]`. Every event `i` starts at `startDayi` and ends at `endDayi`.
+
+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`.
+
+Return *the maximum number of events* you can attend.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2020/02/05/e1.png](https://assets.leetcode.com/uploads/2020/02/05/e1.png)
+
+```
+Input: events = [[1,2],[2,3],[3,4]]
+Output: 3
+Explanation: You can attend all the three events.
+One way to attend them all is as shown.
+Attend the first event on day 1.
+Attend the second event on day 2.
+Attend the third event on day 3.
+
+```
+
+**Example 2:**
+
+```
+Input: events= [[1,2],[2,3],[3,4],[1,2]]
+Output: 4
+
+```
+
+**Example 3:**
+
+```
+Input: events = [[1,4],[4,4],[2,2],[3,4],[1,1]]
+Output: 4
+
+```
+
+**Example 4:**
+
+```
+Input: events = [[1,100000]]
+Output: 1
+
+```
+
+**Example 5:**
+
+```
+Input: events = [[1,1],[1,2],[1,3],[1,4],[1,5],[1,6],[1,7]]
+Output: 7
+
+```
+
+**Constraints:**
+
+- `1 <= events.length <= 10^5`
+- `events[i].length == 2`
+- `1 <= startDayi <= endDayi <= 10^5`
+
+## 题目大意
+
+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。
+
+## 解题思路
+
+- 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。
+- 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。
+- 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。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"sort"
+)
+
+func maxEvents(events [][]int) int {
+	sort.Slice(events, func(i, j int) bool {
+		if events[i][0] == events[j][0] {
+			return events[i][1] < events[j][1]
+		}
+		return events[i][0] < events[j][0]
+	})
+	attended, current := 1, events[0]
+	for i := 1; i < len(events); i++ {
+		prev, event := events[i-1], events[i]
+		if event[0] == prev[0] && event[1] == prev[1] && event[1] == event[0] {
+			continue
+		}
+		start, end := max(current[0], event[0]-1), max(current[1], event[1])
+		if end-start > 0 {
+			current[0] = start + 1
+			current[1] = end
+			attended++
+		}
+	}
+	return attended
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+```