update: 添加问题“2054.两个最好的不重叠活动”的代码和题解 (#1276)

2054: AC.cpp (#1275) - AC,85.19%,53.70%

Signed-off-by: LetMeFly666 <[email protected]>

Author: LetMeFly666

README.md

@@ -696,6 +696,7 @@
 |2044.统计按位或能得到最大值的子集数目|中等|<a href="https://leetcode.cn/problems/count-number-of-maximum-bitwise-or-subsets/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/07/28/LeetCode%202044.%E7%BB%9F%E8%AE%A1%E6%8C%89%E4%BD%8D%E6%88%96%E8%83%BD%E5%BE%97%E5%88%B0%E6%9C%80%E5%A4%A7%E5%80%BC%E7%9A%84%E5%AD%90%E9%9B%86%E6%95%B0%E7%9B%AE/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/149725900" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/count-number-of-maximum-bitwise-or-subsets/solutions/3736072/letmefly-2044tong-ji-an-wei-huo-neng-de-97g9f/" target="_blank">LeetCode题解</a>|
 |2048.下一个更大的数值平衡数|中等|<a href="https://leetcode.cn/problems/next-greater-numerically-balanced-number/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2023/12/09/LeetCode%202048.%E4%B8%8B%E4%B8%80%E4%B8%AA%E6%9B%B4%E5%A4%A7%E7%9A%84%E6%95%B0%E5%80%BC%E5%B9%B3%E8%A1%A1%E6%95%B0/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/134900679" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/next-greater-numerically-balanced-number/solutions/2560353/letmefly-2048xia-yi-ge-geng-da-de-shu-zh-7p4t/" target="_blank">LeetCode题解</a>|
 |2050.并行课程III|困难|<a href="https://leetcode.cn/problems/parallel-courses-iii/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2023/07/28/LeetCode%202050.%E5%B9%B6%E8%A1%8C%E8%AF%BE%E7%A8%8BIII/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/131973511" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/parallel-courses-iii/solutions/2362142/letmefly-2050bing-xing-ke-cheng-iiidfs-b-9tuc/" target="_blank">LeetCode题解</a>|
+|2054.两个最好的不重叠活动|中等|<a href="https://leetcode.cn/problems/two-best-non-overlapping-events/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/12/23/LeetCode%202054.%E4%B8%A4%E4%B8%AA%E6%9C%80%E5%A5%BD%E7%9A%84%E4%B8%8D%E9%87%8D%E5%8F%A0%E6%B4%BB%E5%8A%A8/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/156207232" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/two-best-non-overlapping-events/solutions/3864227/letmefly-2054liang-ge-zui-hao-de-bu-zhon-37h9/" target="_blank">LeetCode题解</a>|
 |2065.最大化一张图中的路径价值|困难|<a href="https://leetcode.cn/problems/maximum-path-quality-of-a-graph/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2024/07/01/LeetCode%202065.%E6%9C%80%E5%A4%A7%E5%8C%96%E4%B8%80%E5%BC%A0%E5%9B%BE%E4%B8%AD%E7%9A%84%E8%B7%AF%E5%BE%84%E4%BB%B7%E5%80%BC/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/140101479" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/maximum-path-quality-of-a-graph/solutions/2827788/letmefly-2065zui-da-hua-yi-zhang-tu-zhon-vvkl/" target="_blank">LeetCode题解</a>|
 |2070.每一个查询的最大美丽值|中等|<a href="https://leetcode.cn/problems/most-beautiful-item-for-each-query/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/03/09/LeetCode%202070.%E6%AF%8F%E4%B8%80%E4%B8%AA%E6%9F%A5%E8%AF%A2%E7%9A%84%E6%9C%80%E5%A4%A7%E7%BE%8E%E4%B8%BD%E5%80%BC/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/146132104" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/most-beautiful-item-for-each-query/solutions/3603183/letmefly-2070mei-yi-ge-cha-xun-de-zui-da-brpw/" target="_blank">LeetCode题解</a>|
 |2073.买票需要的时间|简单|<a href="https://leetcode.cn/problems/time-needed-to-buy-tickets/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2024/10/03/LeetCode%202073.%E4%B9%B0%E7%A5%A8%E9%9C%80%E8%A6%81%E7%9A%84%E6%97%B6%E9%97%B4/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/142691600" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/time-needed-to-buy-tickets/solutions/2937796/letmefly-2073mai-piao-xu-yao-de-shi-jian-i1pt/" target="_blank">LeetCode题解</a>|

Solutions/LeetCode 2054.两个最好的不重叠活动.md

@@ -1,11 +1,12 @@
 ---
-title: 2054.两个最好的不重叠活动：
-date: 2025-12-23 19:00:09
-tags: [题解, LeetCode, 中等, 数组, 二分查找, 动态规划, 排序, 堆（优先队列）]
+title: 2054.两个最好的不重叠活动：二分查找
+date: 2025-12-23 23:45:19
+tags: [题解, LeetCode, 中等, 数组, 二分查找, 动态规划, 排序]
 categories: [题解, LeetCode]
+index_img: https://assets.leetcode.com/uploads/2021/09/21/picture5.png
 ---
 
-# 【LetMeFly】2054.两个最好的不重叠活动：
+# 【LetMeFly】2054.两个最好的不重叠活动：二分查找
 
 力扣题目链接：[https://leetcode.cn/problems/two-best-non-overlapping-events/](https://leetcode.cn/problems/two-best-non-overlapping-events/)
 
@@ -56,28 +57,27 @@ categories: [题解, LeetCode]
 
 
 
-## 解题方法：xx
+## 解题方法：二分查找
 
-11111
+如果只能选一个event，那么好说，哪个价值大选哪个；如果一定要选两个event，假设第二个event选事件e，那么第一个event一定要选结束时间早于e开始时间的所有事件中价值最大的那个。
 
-+ 时间复杂度$O(N^2)$
-+ 空间复杂度$O(N\log N)$
+很显然，为了枚举第一个event的可选范围，可以以结束时间为依据对所有event按从小到大排个序。
+
+接着使用一个（有序）数组maxValue，数组中存放的内容是：到xx时刻为止，单个event的最大价值是多少。排序依据是结束时间。
+
+遍历所有事件，对于某事件e，二分查找maxValue中小于e开始时间中最大的那个，其值加上e的价值即为第二个event选e情况下的最优解。之后更新e结束时间的单个事件最大值。
+
++ 时间复杂度$O(n\log n)$，其中$n=len(events)$
++ 空间复杂度$O(n)$
 
 ### AC代码
 
 #### C++
 
 ```cpp
 /*
- * @Author: LetMeFly
- * @Date: 2025-12-23 13:34:22
- * @LastEditors: LetMeFly.xyz
  * @LastEditTime: 2025-12-23 18:58:01
  */
-#if defined(_WIN32) || defined(__APPLE__)
-#include "_[1,2]toVector.h"
-#endif
-
 class Solution {
 public:
     int maxTwoEvents(vector<vector<int>>& events) {
@@ -101,6 +101,6 @@ public:
 };
 ```
 
-> 同步发文于[CSDN](https://letmefly.blog.csdn.net/article/details/--------------------------)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/2025/12/23/LeetCode%202054.%E4%B8%A4%E4%B8%AA%E6%9C%80%E5%A5%BD%E7%9A%84%E4%B8%8D%E9%87%8D%E5%8F%A0%E6%B4%BB%E5%8A%A8/)哦~
+> 同步发文于[CSDN](https://letmefly.blog.csdn.net/article/details/156207232)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/2025/12/23/LeetCode%202054.%E4%B8%A4%E4%B8%AA%E6%9C%80%E5%A5%BD%E7%9A%84%E4%B8%8D%E9%87%8D%E5%8F%A0%E6%B4%BB%E5%8A%A8/)哦~
 >
 > 千篇源码题解[已开源](https://github.com/LetMeFly666/LeetCode)

log.py

@@ -0,0 +1,53 @@
+'''
+Author: LetMeFly
+Date: 2025-12-23 22:37:57
+LastEditors: LetMeFly.xyz
+LastEditTime: 2025-12-23 22:43:45
+'''
+
+import re
+from typing import Optional
+from datetime import datetime
+
+INPUT_FILE = "C:/Users/LetMe/Desktop/let_proc_snapshot.log"
+OUTPUT_FILE = "C:/Users/LetMe/Desktop/proc_snapshot_5min.log"
+
+# 匹配头部时间行
+HEADER_RE = re.compile(r"^===== (.+) =====$")
+
+def parse_time(line: str) -> Optional[datetime]:
+    m = HEADER_RE.match(line.strip())
+    if not m:
+        return None
+    return datetime.strptime(m.group(1), "%Y-%m-%d %H:%M:%S")
+
+def should_keep(ts: datetime) -> bool:
+    return ts.minute % 5 == 0
+
+def main():
+    with open(INPUT_FILE, "r", encoding="utf-8") as fin, \
+         open(OUTPUT_FILE, "w", encoding="utf-8") as fout:
+
+        block_lines = []
+        block_time = None
+
+        for line in fin:
+            ts = parse_time(line)
+            if ts:
+                # 新 block 开始，先处理上一个
+                if block_time and should_keep(block_time):
+                    fout.writelines(block_lines)
+
+                block_lines = [line]
+                block_time = ts
+            else:
+                block_lines.append(line)
+
+        # 文件结束，别忘了最后一个 block
+        if block_time and should_keep(block_time):
+            fout.writelines(block_lines)
+
+    print(f"Resample done: {OUTPUT_FILE}")
+
+if __name__ == "__main__":
+    main()

todo

@@ -1,2 +1,54 @@
-daka
-write a script to delete origin/copilot*
+看下2054最优解 看下copilot评论
+write a script to delete origin/copilot*
+article - 服务器内存泄漏分析  pkill -f "manage.py crontab run"
+
+豆包搜文章
+cdn.letmefly.xyz/img/blog/DoubaoChatSearch/whenEnter.png
+how2use.jpg
+chatResult.jpg
+{
+    "pluginId": "Send_Message",
+    "payload": {
+        "text": "letleet blog到底有多少宝藏"
+    }
+}
+
+{
+    "pluginId": "Send_Message",
+    "payload": {
+        "text": "letleet blog到底有多少宝藏",
+        "reportParams": {
+            "content_source": "paste",
+            "content_source_template_id": "",
+            "send_source": "guidance_page",
+            "scene": "keyboard",
+            "is_long_file": 0,
+            "new_guidance": "1",
+            "is_from_screen_shot": 0,
+            "is_from_new_tab": 1
+        },
+        "conversationId": "local_61827421334957",
+        "extraExt": {
+            "input_skill": "{\"skill_type\":20}",
+            "use_deep_think": "0"
+        },
+        "loginSource": "new_tab",
+        "startJumpTime": 1766473291953
+    },
+    "pfTrackParams": {
+        "startJumpTime": 1766473291953,
+        "phase1EventName": "pc_new_tab_chat_open_jump_phase1_duration",
+        "phase2EventName": "pc_new_tab_chat_open_jump_phase2_duration",
+        "phase3EventName": "pc_new_tab_chat_open_jump_phase3_duration",
+        "phaseNavEventName": "pc_new_tab_chat_open_jump_phasenav_duration",
+        "phaseChatRenderEventName": "pc_new_tab_chat_open_jump_chatrender_duration"
+    },
+    "options": {
+        "deepThinkingActiveType": "0",
+        "superTaskStatus": {
+            "switchValue": 0,
+            "taskMode": 0,
+            "safeToken": ""
+        }
+    }
+}