Merge pull request #696 from LetMeFly666/2264

添加问题“2264.字符串中最大的3位相同数字”的代码和题解

Author: LetMeFly666

Codes/2264-largest-3-same-digit-number-in-string.cpp

@@ -0,0 +1,41 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-01-08 15:25:00
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-01-08 15:33:22
+ */
+#ifdef _WIN32
+#include "_[1,2]toVector.h"
+#endif
+
+#ifdef FirstTry  // WA - 必须是子字符串
+class Solution {
+public:
+    string largestGoodInteger(string num) {
+        int bin[10] = {0};
+        for (char c : num) {
+            bin[c - '0']++;
+        }
+        for (int i = 9; i >= 0; i--) {
+            if (bin[i] >= 3) {
+                return to_string(i) + to_string(i) + to_string(i);
+            }
+        }
+        return "";
+    }
+};
+#else  // FirstTry
+// SecondTry
+class Solution {
+public:
+    string largestGoodInteger(string& num) {
+        char M = '/';  // ASCII在0前一个
+        for (int i = 2; i < num.size(); i++) {
+            if (num[i] == num[i - 1] && num[i] == num[i - 2]) {
+                M = max(M, num[i]);
+            }
+        }
+        return M == '/' ? string() : string(3, M);
+    }
+};
+#endif  // FirstTry

Codes/2264-largest-3-same-digit-number-in-string.go

@@ -0,0 +1,20 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-01-08 15:43:44
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-01-08 15:46:23
+ */
+package main
+
+func largestGoodInteger(num string) string {
+    M := byte('/')
+    for i := 2; i < len(num); i++ {
+        if num[i] > M && num[i] == num[i - 1] && num[i] == num[i - 2] {
+            M = num[i]
+        }
+    }
+    if M == '/' {
+        return ""
+    }
+    return string(M) + string(M) + string(M)
+}

Codes/2264-largest-3-same-digit-number-in-string.java

@@ -0,0 +1,17 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-01-08 15:37:06
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-01-08 15:41:37
+ */
+class Solution {
+    public String largestGoodInteger(String num) {
+        char M = '/';
+        for (int i = 2; i < num.length(); i++) {
+            if (num.charAt(i) == num.charAt(i - 1) && num.charAt(i) == num.charAt(i - 2)) {
+                M = (char) Math.max(M, num.charAt(i));
+            }
+        }
+        return M == '/' ? "" : "" + M + M + M;
+    }
+}

Codes/2264-largest-3-same-digit-number-in-string.py

@@ -0,0 +1,13 @@
+'''
+Author: LetMeFly
+Date: 2025-01-08 15:34:27
+LastEditors: LetMeFly.xyz
+LastEditTime: 2025-01-08 15:36:15
+'''
+class Solution:
+    def largestGoodInteger(self, s: str) -> str:
+        M = '/'
+        for i in range(2, len(s)):
+            if s[i] == s[i - 1] == s[i - 2]:
+                M = max(M, s[i])
+        return '' if M == '/' else M * 3

README.md

@@ -538,6 +538,7 @@
 |2244.完成所有任务需要的最少轮数|中等|<a href="https://leetcode.cn/problems/minimum-rounds-to-complete-all-tasks/solutions/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2024/05/14/LeetCode%202244.%E5%AE%8C%E6%88%90%E6%89%80%E6%9C%89%E4%BB%BB%E5%8A%A1%E9%9C%80%E8%A6%81%E7%9A%84%E6%9C%80%E5%B0%91%E8%BD%AE%E6%95%B0/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/138849002" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/minimum-rounds-to-complete-all-tasks/solutions/2777174/letmefly-2244wan-cheng-suo-you-ren-wu-xu-n7bh/" target="_blank">LeetCode题解</a>|
 |2251.花期内花的数目|困难|<a href="https://leetcode.cn/problems/number-of-flowers-in-full-bloom/solutions/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2023/09/28/LeetCode%202251.%E8%8A%B1%E6%9C%9F%E5%86%85%E8%8A%B1%E7%9A%84%E6%95%B0%E7%9B%AE/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/133378624" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/number-of-flowers-in-full-bloom/solutions/2462220/letmefly-2251hua-qi-nei-hua-de-shu-mu-pa-nfvo/" target="_blank">LeetCode题解</a>|
 |2258.逃离火灾|困难|<a href="https://leetcode.cn/problems/escape-the-spreading-fire/solutions/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2023/11/09/LeetCode%202258.%E9%80%83%E7%A6%BB%E7%81%AB%E7%81%BE/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/134331955" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/escape-the-spreading-fire/solutions/2520786/letmefly-2258tao-chi-huo-zai-yan-du-you-kj2r4/" target="_blank">LeetCode题解</a>|
+|2264.字符串中最大的3位相同数字|简单|<a href="https://leetcode.cn/problems/largest-3-same-digit-number-in-string/solutions/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/01/08/LeetCode%202264.%E5%AD%97%E7%AC%A6%E4%B8%B2%E4%B8%AD%E6%9C%80%E5%A4%A7%E7%9A%843%E4%BD%8D%E7%9B%B8%E5%90%8C%E6%95%B0%E5%AD%97/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/145011261" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/largest-3-same-digit-number-in-string/solutions/3042054/letmefly-2264zi-fu-chuan-zhong-zui-da-de-5vja/" target="_blank">LeetCode题解</a>|
 |2274.不含特殊楼层的最大连续楼层数|中等|<a href="https://leetcode.cn/problems/maximum-consecutive-floors-without-special-floors/solutions/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/01/06/LeetCode%202274.%E4%B8%8D%E5%90%AB%E7%89%B9%E6%AE%8A%E6%A5%BC%E5%B1%82%E7%9A%84%E6%9C%80%E5%A4%A7%E8%BF%9E%E7%BB%AD%E6%A5%BC%E5%B1%82%E6%95%B0/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/144972086" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/maximum-consecutive-floors-without-special-floors/solutions/3040396/letmefly-2274bu-han-te-shu-lou-ceng-de-z-321m/" target="_blank">LeetCode题解</a>|
 |2276.统计区间中的整数数目|困难|<a href="https://leetcode.cn/problems/count-integers-in-intervals/solutions/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2023/12/16/LeetCode%202276.%E7%BB%9F%E8%AE%A1%E5%8C%BA%E9%97%B4%E4%B8%AD%E7%9A%84%E6%95%B4%E6%95%B0%E6%95%B0%E7%9B%AE/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/135036679" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/count-integers-in-intervals/solutions/2568803/letmefly-2276tong-ji-qu-jian-zhong-de-zh-y3i6/" target="_blank">LeetCode题解</a>|
 |2283.判断一个数的数字计数是否等于数位的值|简单|<a href="https://leetcode.cn/problems/check-if-number-has-equal-digit-count-and-digit-value/solutions/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2023/01/11/LeetCode%202283.%E5%88%A4%E6%96%AD%E4%B8%80%E4%B8%AA%E6%95%B0%E7%9A%84%E6%95%B0%E5%AD%97%E8%AE%A1%E6%95%B0%E6%98%AF%E5%90%A6%E7%AD%89%E4%BA%8E%E6%95%B0%E4%BD%8D%E7%9A%84%E5%80%BC/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/128652351" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/check-if-number-has-equal-digit-count-and-digit-value/solutions/2057234/letmefly-2283pan-duan-yi-ge-shu-de-shu-z-4olo/" target="_blank">LeetCode题解</a>|

Solutions/LeetCode 2264.字符串中最大的3位相同数字.md

@@ -0,0 +1,164 @@
+---
+title: 2264.字符串中最大的 3 位相同数字
+date: 2025-01-08 15:48:37
+tags: [题解, LeetCode, 简单, 字符串]
+---
+
+# 【LetMeFly】2264.字符串中最大的 3 位相同数字：遍历
+
+力扣题目链接：[https://leetcode.cn/problems/largest-3-same-digit-number-in-string/](https://leetcode.cn/problems/largest-3-same-digit-number-in-string/)
+
+<p>给你一个字符串 <code>num</code> ，表示一个大整数。如果一个整数满足下述所有条件，则认为该整数是一个 <strong>优质整数</strong> ：</p>
+
+<ul>
+	<li>该整数是 <code>num</code> 的一个长度为 <code>3</code> 的 <strong>子字符串</strong> 。</li>
+	<li>该整数由唯一一个数字重复 <code>3</code> 次组成。</li>
+</ul>
+
+<p>以字符串形式返回 <strong>最大的优质整数</strong> 。如果不存在满足要求的整数，则返回一个空字符串 <code>""</code> 。</p>
+
+<p><strong>注意：</strong></p>
+
+<ul>
+	<li><strong>子字符串</strong> 是字符串中的一个连续字符序列。</li>
+	<li><code>num</code> 或优质整数中可能存在 <strong>前导零</strong> 。</li>
+</ul>
+
+<p>&nbsp;</p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre>
+<strong>输入：</strong>num = "6<em><strong>777</strong></em>133339"
+<strong>输出：</strong>"777"
+<strong>解释：</strong>num 中存在两个优质整数："777" 和 "333" 。
+"777" 是最大的那个，所以返回 "777" 。
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre>
+<strong>输入：</strong>num = "23<em><strong>000</strong></em>19"
+<strong>输出：</strong>"000"
+<strong>解释：</strong>"000" 是唯一一个优质整数。
+</pre>
+
+<p><strong>示例 3：</strong></p>
+
+<pre>
+<strong>输入：</strong>num = "42352338"
+<strong>输出：</strong>""
+<strong>解释：</strong>不存在长度为 3 且仅由一个唯一数字组成的整数。因此，不存在优质整数。
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>3 &lt;= num.length &lt;= 1000</code></li>
+	<li><code>num</code> 仅由数字（<code>0</code> - <code>9</code>）组成</li>
+</ul>
+
+
+    
+## 解题方法：遍历
+
+使用一个变量记录最大的优质整数的字符，遍历字符串并不断更新这个最大值即可。
+
++ 时间复杂度$O(len(num))$
++ 空间复杂度$O(1)$
+
+### AC代码
+
+#### C++
+
+```cpp
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-01-08 15:25:00
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-01-08 15:33:22
+ */
+class Solution {
+public:
+    string largestGoodInteger(string& num) {
+        char M = '/';  // ASCII在0前一个
+        for (int i = 2; i < num.size(); i++) {
+            if (num[i] == num[i - 1] && num[i] == num[i - 2]) {
+                M = max(M, num[i]);
+            }
+        }
+        return M == '/' ? string() : string(3, M);
+    }
+};
+```
+
+#### Python
+
+```python
+'''
+Author: LetMeFly
+Date: 2025-01-08 15:34:27
+LastEditors: LetMeFly.xyz
+LastEditTime: 2025-01-08 15:36:15
+'''
+class Solution:
+    def largestGoodInteger(self, s: str) -> str:
+        M = '/'
+        for i in range(2, len(s)):
+            if s[i] == s[i - 1] == s[i - 2]:
+                M = max(M, s[i])
+        return '' if M == '/' else M * 3
+```
+
+#### Java
+
+```java
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-01-08 15:37:06
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-01-08 15:41:37
+ */
+class Solution {
+    public String largestGoodInteger(String num) {
+        char M = '/';
+        for (int i = 2; i < num.length(); i++) {
+            if (num.charAt(i) == num.charAt(i - 1) && num.charAt(i) == num.charAt(i - 2)) {
+                M = (char) Math.max(M, num.charAt(i));
+            }
+        }
+        return M == '/' ? "" : "" + M + M + M;
+    }
+}
+```
+
+#### Go
+
+```go
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-01-08 15:43:44
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-01-08 15:46:23
+ */
+package main
+
+func largestGoodInteger(num string) string {
+    M := byte('/')
+    for i := 2; i < len(num); i++ {
+        if num[i] > M && num[i] == num[i - 1] && num[i] == num[i - 2] {
+            M = num[i]
+        }
+    }
+    if M == '/' {
+        return ""
+    }
+    return string(M) + string(M) + string(M)
+}
+```
+
+> 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/2025/01/08/LeetCode%202264.%E5%AD%97%E7%AC%A6%E4%B8%B2%E4%B8%AD%E6%9C%80%E5%A4%A7%E7%9A%843%E4%BD%8D%E7%9B%B8%E5%90%8C%E6%95%B0%E5%AD%97/)哦~
+>
+> Tisfy：[https://letmefly.blog.csdn.net/article/details/145011261](https://letmefly.blog.csdn.net/article/details/145011261)

api/calendar/README.md

@@ -2,7 +2,7 @@
  * @Author: LetMeFly
  * @Date: 2024-12-15 16:10:07
  * @LastEditors: LetMeFly.xyz
- * @LastEditTime: 2025-01-05 13:38:33
+ * @LastEditTime: 2025-01-07 23:43:43
 -->
 # 目的
 
@@ -27,6 +27,9 @@
 # TODO
 
 - [ ] 删除标签时，前端提醒“所有任务的xx标签将会被移除”
+- [ ] 前端，创建完一个事件后创建另一个事件时内容不清空
+- [ ] 前端，默认显示时间是UTC时间而不是本地时间
+- [ ] 后端，taskTag数据库中删除一条记录时，这个taskId的所有taskTag记录都会被删除
 - [ ] 前端，事件修改 - 这个可以全部删了重建
 - [ ] 后端 - 事件创建 - 时间覆盖重叠问题
 - [ ] fix: 前端 - 拖拽后默认时间不正确
@@ -878,13 +881,61 @@ html表格，设置所有列等宽。
 创建事件时，可以输入事件标题、事件描述，默认依据用户的拖拽范围给定一个起止时间，用户也可以调整起止时间。
 
 <hr/>
+
+我有一个Calendar_Tasks数据表，里面有task信息和userid信息；
+我有一个Calendar_TaskTag数据表，里面有taskId和tagId的对应信息。
+
+我想通过userid查询出这个user的所有任务，以及每个任务对应的tagId。我应该如何查询？
+
+```
+```
+
 <hr/>
+
+解释left join
+
 <hr/>
+
+这样查询的效率如何
+
 <hr/>
+
+left join的时间复杂度是多少？
+
 <hr/>
+
+这样左表中的信息是不是会一个tag重复一次
+
 <hr/>
+
+GROUP_CONCAT这个好，请详细解释之
+
 <hr/>
+
+如果task表列比较多的话，可否Select task.*
+
 <hr/>
+
+这样会每个userid都关联一次，我可用做到只管理指定userid的task吗？
+
+我尝试如下代码报错了：
+
+```
+SELECT
+    Calendar_Tasks.*,
+    GROUP_CONCAT(Calendar_TaskTag.tagId) as tagIds
+FROM
+    Calendar_Tasks
+LEFT JOIN
+    Calendar_TaskTag
+ON
+    Calendar_Tasks.taskId = Calendar_TaskTag.taskId
+GROUP BY
+    Calendar_Tasks.taskId
+WHERE
+    Calendar_Tasks.userid = 1
+```
+
 <hr/>
 <hr/>
 <hr/>

api/calendar/back/README.md

@@ -2,7 +2,7 @@
  * @Author: LetMeFly
  * @Date: 2024-12-20 22:12:40
  * @LastEditors: LetMeFly.xyz
- * @LastEditTime: 2024-12-24 10:43:09
+ * @LastEditTime: 2025-01-07 23:47:31
 -->
 # calendar后端
 
@@ -85,15 +85,17 @@ None。 TODO: 起止时间
         "description": "Let Calendar开发",
         "startTime": "2024-12-17T10:00:00",
         "during": 60,
-        "userid": 1
+        "userid": 1,
+        "tagIds": null
     },
     {
-        "taskId": 13,
-        "title": "开发",
-        "description": "LetCalendar开发",
-        "startTime": "2024-12-20T10:00:00",
-        "during": 60,
-        "userid": 1
+        "taskId": 18,
+        "title": "聊天",
+        "description": "哄对象不生气[trumble]",
+        "startTime": "2025-01-07T12:20:00",
+        "during": 180,
+        "userid": 1,
+        "tagIds": "5,7"
     }
 ]
 ```