Merge pull request #684 from LetMeFly666/729

添加问题“729.我的日程安排表I”的代码和题解

Author: LetMeFly666

Codes/0729-my-calendar-i.cpp

@@ -0,0 +1,35 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-01-02 16:13:40
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-01-02 16:15:23
+ */
+#ifdef _WIN32
+#include "_[1,2]toVector.h"
+#endif
+
+class MyCalendar {
+private:
+    vector<int> start, end;
+public:
+    MyCalendar() {
+        
+    }
+    
+    bool book(int startTime, int endTime) {
+        for (int i = 0; i < start.size(); i++) {
+            if (!(end[i] <= startTime || endTime <= start[i])) {
+                return false;
+            }
+        }
+        start.push_back(startTime);
+        end.push_back(endTime);
+        return true;
+    }
+};
+
+/**
+ * Your MyCalendar object will be instantiated and called as such:
+ * MyCalendar* obj = new MyCalendar();
+ * bool param_1 = obj->book(startTime,endTime);
+ */

Codes/0729-my-calendar-i.go

@@ -0,0 +1,33 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-01-02 16:27:36
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-01-02 16:32:36
+ */
+package main
+
+type pair struct{ start, end int }
+type MyCalendar []pair
+
+
+func Constructor() MyCalendar {
+    return MyCalendar{}
+}
+
+
+func (this *MyCalendar) Book(startTime int, endTime int) bool {
+    for _, p := range *this {
+        if endTime > p.start && p.end > startTime {
+            return false
+        }
+    }
+    *this = append(*this, pair{startTime, endTime})
+    return true
+}
+
+
+/**
+ * Your MyCalendar object will be instantiated and called as such:
+ * obj := Constructor();
+ * param_1 := obj.Book(startTime,endTime);
+ */

Codes/0729-my-calendar-i.java

@@ -0,0 +1,33 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-01-02 16:22:47
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-01-02 16:27:07
+ */
+import java.util.ArrayList;
+
+class MyCalendar {
+    private ArrayList<Integer> start, end;
+
+    public MyCalendar() {
+        start = new ArrayList<Integer>();
+        end = new ArrayList<Integer>();
+    }
+    
+    public boolean book(int startTime, int endTime) {
+        for (int i = 0; i < start.size(); i++) {
+            if (end.get(i) > startTime && endTime > start.get(i)) {
+                return false;
+            }
+        }
+        start.add(startTime);
+        end.add(endTime);
+        return true;
+    }
+}
+
+/**
+ * Your MyCalendar object will be instantiated and called as such:
+ * MyCalendar obj = new MyCalendar();
+ * boolean param_1 = obj.book(startTime,endTime);
+ */

Codes/0729-my-calendar-i.py

@@ -0,0 +1,22 @@
+'''
+Author: LetMeFly
+Date: 2025-01-02 16:16:58
+LastEditors: LetMeFly.xyz
+LastEditTime: 2025-01-02 16:22:15
+'''
+# 产生交集：!(e1 <= s2 || e2 <= s1) 等价于 e1 > s2 && e2 > s1
+class MyCalendar:
+
+    def __init__(self):
+        self.bookList = []
+
+    def book(self, startTime: int, endTime: int) -> bool:
+        if any(e > startTime and endTime > s for s, e in self.bookList):
+            return False
+        self.bookList.append((startTime, endTime))
+        return True
+
+
+# Your MyCalendar object will be instantiated and called as such:
+# obj = MyCalendar()
+# param_1 = obj.book(startTime,endTime)

README.md

@@ -252,6 +252,7 @@
 |0714.买卖股票的最佳时机含手续费|中等|<a href="https://leetcode.cn/problems/best-time-to-buy-and-sell-stock-with-transaction-fee/solutions/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2023/10/06/LeetCode%200714.%E4%B9%B0%E5%8D%96%E8%82%A1%E7%A5%A8%E7%9A%84%E6%9C%80%E4%BD%B3%E6%97%B6%E6%9C%BA%E5%90%AB%E6%89%8B%E7%BB%AD%E8%B4%B9/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/133609633" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/best-time-to-buy-and-sell-stock-with-transaction-fee/solutions/2469756/letmefly-714mai-mai-gu-piao-de-zui-jia-s-t228/" target="_blank">LeetCode题解</a>|
 |0722.删除注释|中等|<a href="https://leetcode.cn/problems/remove-comments/solutions/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2023/08/03/LeetCode%200722.%E5%88%A0%E9%99%A4%E6%B3%A8%E9%87%8A/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/132075300" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/remove-comments/solutions/2370633/letmefly-722shan-chu-zhu-shi-by-tisfy-hbzi/" target="_blank">LeetCode题解</a>|
 |0724.寻找数组的中心下标|简单|<a href="https://leetcode.cn/problems/find-pivot-index/solutions/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2024/07/08/LeetCode%200724.%E5%AF%BB%E6%89%BE%E6%95%B0%E7%BB%84%E7%9A%84%E4%B8%AD%E5%BF%83%E4%B8%8B%E6%A0%87/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/140266165" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/find-pivot-index/solutions/2835007/letmefly-724xun-zhao-shu-zu-de-zhong-xin-36y4/" target="_blank">LeetCode题解</a>|
+|0729.我的日程安排表I|中等|<a href="https://leetcode.cn/problems/my-calendar-i/solutions/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/01/02/LeetCode%200729.%E6%88%91%E7%9A%84%E6%97%A5%E7%A8%8B%E5%AE%89%E6%8E%92%E8%A1%A8I/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/144889921" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/my-calendar-i/solutions/3036765/letmefly-729wo-de-ri-cheng-an-pai-biao-i-5yzd/" target="_blank">LeetCode题解</a>|
 |0735.行星碰撞|中等|<a href="https://leetcode.cn/problems/asteroid-collision/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2022/07/13/LeetCode%200735.%E8%A1%8C%E6%98%9F%E7%A2%B0%E6%92%9E/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/125774687" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/asteroid-collision/solution/by-tisfy-nxaa/" target="_blank">LeetCode题解</a>|
 |0746.使用最小花费爬楼梯|简单|<a href="https://leetcode.cn/problems/min-cost-climbing-stairs/solutions/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2023/12/17/LeetCode%200746.%E4%BD%BF%E7%94%A8%E6%9C%80%E5%B0%8F%E8%8A%B1%E8%B4%B9%E7%88%AC%E6%A5%BC%E6%A2%AF/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/135046961" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/min-cost-climbing-stairs/solutions/2569770/letmefly-746shi-yong-zui-xiao-hua-fei-pa-pz2r/" target="_blank">LeetCode题解</a>|
 |0749.隔离病毒|困难|<a href="https://leetcode.cn/problems/contain-virus/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2022/07/18/LeetCode%200749.%E9%9A%94%E7%A6%BB%E7%97%85%E6%AF%92/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/125846470" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/contain-virus/solution/letmefly-749ge-chi-bing-du-by-tisfy-746u/" target="_blank">LeetCode题解</a>|

Solutions/LeetCode 0729.我的日程安排表I.md

@@ -0,0 +1,212 @@
+---
+title: 729.我的日程安排表 I
+date: 2025-01-02 16:33:43
+tags: [题解, LeetCode, 中等, 设计, 线段树, 数组, 二分查找, 有序集合, 模拟]
+---
+
+# 【LetMeFly】729.我的日程安排表 I：既然是I，那就暴力模拟
+
+力扣题目链接：[https://leetcode.cn/problems/my-calendar-i/](https://leetcode.cn/problems/my-calendar-i/)
+
+<p>实现一个 <code>MyCalendar</code> 类来存放你的日程安排。如果要添加的日程安排不会造成 <strong>重复预订</strong> ，则可以存储这个新的日程安排。</p>
+
+<p>当两个日程安排有一些时间上的交叉时（例如两个日程安排都在同一时间内），就会产生 <strong>重复预订</strong> 。</p>
+
+<p>日程可以用一对整数 <code>startTime</code> 和 <code>endTime</code> 表示，这里的时间是半开区间，即 <code>[startTime, endTime)</code>, 实数&nbsp;<code>x</code> 的范围为， &nbsp;<code>startTime &lt;= x &lt; endTime</code> 。</p>
+
+<p>实现 <code>MyCalendar</code> 类：</p>
+
+<ul>
+	<li><code>MyCalendar()</code> 初始化日历对象。</li>
+	<li><code>boolean book(int startTime, int endTime)</code> 如果可以将日程安排成功添加到日历中而不会导致重复预订，返回 <code>true</code> 。否则，返回 <code>false</code>&nbsp;并且不要将该日程安排添加到日历中。</li>
+</ul>
+
+<p>&nbsp;</p>
+
+<p><strong class="example">示例：</strong></p>
+
+<pre>
+<strong>输入：</strong>
+["MyCalendar", "book", "book", "book"]
+[[], [10, 20], [15, 25], [20, 30]]
+<strong>输出：</strong>
+[null, true, false, true]
+
+<strong>解释：</strong>
+MyCalendar myCalendar = new MyCalendar();
+myCalendar.book(10, 20); // return True
+myCalendar.book(15, 25); // return False ，这个日程安排不能添加到日历中，因为时间 15 已经被另一个日程安排预订了。
+myCalendar.book(20, 30); // return True ，这个日程安排可以添加到日历中，因为第一个日程安排预订的每个时间都小于 20 ，且不包含时间 20 。</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>0 &lt;= start &lt; end &lt;= 10<sup>9</sup></code></li>
+	<li>每个测试用例，调用 <code>book</code> 方法的次数最多不超过 <code>1000</code> 次。</li>
+</ul>
+
+
+    
+## 解题方法：模拟
+
+这道题是“日程安排**I**”，数据量是$10^3$量级的，$O(n^2)$时间复杂度的算法就能过。因此直接模拟就好了。
+
+使用一个数组或两个数组记录每次成功被预定的起止时间，对于一个新的预定，如果与历史上每个起止时间都不冲突，则记入预定数组并返回`true`；否则返回`false`。
+
+对于起止时间是`[s1, e1)`和`[s2, e2)`两个时间段，如何判断二者是否有冲突？
+
+> 如果`e1 > s2 且 e2 > s1`则有冲突。
+>
+> 如果不理解，也可以使用下面是思路逆向推导：
+>
+> 两端时间没有交集说明`s1 >= e2 或 s2 >= e1`，对其取非就是了。
+
++ 时间复杂度$O(n^2)$，其中$n$是调用总次数
++ 空间复杂度$O(n)$
+
+### AC代码
+
+#### C++
+
+```cpp
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-01-02 16:13:40
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-01-02 16:15:23
+ */
+class MyCalendar {
+private:
+    vector<int> start, end;
+public:
+    MyCalendar() {
+        
+    }
+    
+    bool book(int startTime, int endTime) {
+        for (int i = 0; i < start.size(); i++) {
+            if (!(end[i] <= startTime || endTime <= start[i])) {
+                return false;
+            }
+        }
+        start.push_back(startTime);
+        end.push_back(endTime);
+        return true;
+    }
+};
+
+/**
+ * Your MyCalendar object will be instantiated and called as such:
+ * MyCalendar* obj = new MyCalendar();
+ * bool param_1 = obj->book(startTime,endTime);
+ */
+```
+
+#### Python
+
+```python
+'''
+Author: LetMeFly
+Date: 2025-01-02 16:16:58
+LastEditors: LetMeFly.xyz
+LastEditTime: 2025-01-02 16:22:15
+'''
+# 产生交集：!(e1 <= s2 || e2 <= s1) 等价于 e1 > s2 && e2 > s1
+class MyCalendar:
+
+    def __init__(self):
+        self.bookList = []
+
+    def book(self, startTime: int, endTime: int) -> bool:
+        if any(e > startTime and endTime > s for s, e in self.bookList):
+            return False
+        self.bookList.append((startTime, endTime))
+        return True
+
+
+# Your MyCalendar object will be instantiated and called as such:
+# obj = MyCalendar()
+# param_1 = obj.book(startTime,endTime)
+```
+
+#### Java
+
+```java
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-01-02 16:22:47
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-01-02 16:27:07
+ */
+import java.util.ArrayList;
+
+class MyCalendar {
+    private ArrayList<Integer> start, end;
+
+    public MyCalendar() {
+        start = new ArrayList<Integer>();
+        end = new ArrayList<Integer>();
+    }
+    
+    public boolean book(int startTime, int endTime) {
+        for (int i = 0; i < start.size(); i++) {
+            if (end.get(i) > startTime && endTime > start.get(i)) {
+                return false;
+            }
+        }
+        start.add(startTime);
+        end.add(endTime);
+        return true;
+    }
+}
+
+/**
+ * Your MyCalendar object will be instantiated and called as such:
+ * MyCalendar obj = new MyCalendar();
+ * boolean param_1 = obj.book(startTime,endTime);
+ */
+```
+
+#### Go
+
+```go
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-01-02 16:27:36
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-01-02 16:32:36
+ */
+package main
+
+type pair struct{ start, end int }
+type MyCalendar []pair
+
+
+func Constructor() MyCalendar {
+    return MyCalendar{}
+}
+
+
+func (this *MyCalendar) Book(startTime int, endTime int) bool {
+    for _, p := range *this {
+        if endTime > p.start && p.end > startTime {
+            return false
+        }
+    }
+    *this = append(*this, pair{startTime, endTime})
+    return true
+}
+
+
+/**
+ * Your MyCalendar object will be instantiated and called as such:
+ * obj := Constructor();
+ * param_1 := obj.Book(startTime,endTime);
+ */
+```
+
+> 同步发文于CSDN和我的[个人博客](https://blog.letmefly.xyz/)，原创不易，转载经作者同意后请附上[原文链接](https://blog.letmefly.xyz/2025/01/02/LeetCode%200729.%E6%88%91%E7%9A%84%E6%97%A5%E7%A8%8B%E5%AE%89%E6%8E%92%E8%A1%A8I/)哦~
+>
+> Tisfy：[https://letmefly.blog.csdn.net/article/details/144889921](https://letmefly.blog.csdn.net/article/details/144889921)