Merge branch 'doocs:main' into main

Author: Nothing-avil

lcof2/剑指 Offer II 113. 课程顺序/README.md

@@ -56,19 +56,27 @@
 
 ## 解法
 
-### 方法一
+### 方法一：拓扑排序
+
+拓扑排序的思路是，先统计每个节点的入度，然后从入度为 0 的节点开始，依次删除这些节点，同时更新与这些节点相连的节点的入度，直到所有节点都被删除。
+
+这里使用队列来存储入度为 0 的节点，每次从队列中取出一个节点，将其加入结果数组中，然后遍历与这个节点相连的节点，将这些节点的入度减 1，如果减 1 后入度为 0，则将这些节点加入队列中。
+
+最后判断结果数组的长度是否等于节点的个数，如果等于则返回结果数组，否则返回空数组。
+
+时间复杂度 $O(n + m)$，空间复杂度 $O(n + m)$。其中 $n$ 和 $m$ 分别是节点的个数和边的个数。
 
 <!-- tabs:start -->
 
 ```python
 class Solution:
     def findOrder(self, numCourses: int, prerequisites: List[List[int]]) -> List[int]:
-        g = defaultdict(list)
+        g = [[] for _ in range(numCourses)]
         indeg = [0] * numCourses
         for a, b in prerequisites:
             g[b].append(a)
             indeg[a] += 1
-        q = deque([i for i, v in enumerate(indeg) if v == 0])
+        q = deque(i for i, v in enumerate(indeg) if v == 0)
         ans = []
         while q:
             i = q.popleft()
@@ -117,23 +125,29 @@ class Solution {
 class Solution {
 public:
     vector<int> findOrder(int numCourses, vector<vector<int>>& prerequisites) {
-        vector<vector<int>> g(numCourses);
+        vector<int> g[numCourses];
         vector<int> indeg(numCourses);
         for (auto& p : prerequisites) {
             int a = p[0], b = p[1];
             g[b].push_back(a);
             ++indeg[a];
         }
         queue<int> q;
-        for (int i = 0; i < numCourses; ++i)
-            if (indeg[i] == 0) q.push(i);
+        for (int i = 0; i < numCourses; ++i) {
+            if (indeg[i] == 0) {
+                q.push(i);
+            }
+        }
         vector<int> ans;
-        while (!q.empty()) {
+        while (q.size()) {
             int i = q.front();
             q.pop();
             ans.push_back(i);
-            for (int j : g[i])
-                if (--indeg[j] == 0) q.push(j);
+            for (int j : g[i]) {
+                if (--indeg[j] == 0) {
+                    q.push(j);
+                }
+            }
         }
         return ans.size() == numCourses ? ans : vector<int>();
     }
@@ -176,61 +190,55 @@ func findOrder(numCourses int, prerequisites [][]int) []int {
 
 ```ts
 function findOrder(numCourses: number, prerequisites: number[][]): number[] {
-    let g = Array.from({ length: numCourses }, () => []);
-    let indeg = new Array(numCourses).fill(0);
-    for (let [a, b] of prerequisites) {
+    const g: number[][] = Array.from({ length: numCourses }, () => []);
+    const indeg: number[] = Array(numCourses).fill(0);
+    for (const [a, b] of prerequisites) {
         g[b].push(a);
         ++indeg[a];
     }
-    let q = [];
-    for (let i = 0; i < numCourses; ++i) {
-        if (!indeg[i]) {
-            q.push(i);
-        }
-    }
-    let ans = [];
+    const q: number[] = indeg.map((v, i) => (v === 0 ? i : -1)).filter(v => v !== -1);
+    const ans: number[] = [];
     while (q.length) {
-        const i = q.shift();
+        const i = q.pop()!;
         ans.push(i);
-        for (let j of g[i]) {
-            if (--indeg[j] == 0) {
+        for (const j of g[i]) {
+            if (--indeg[j] === 0) {
                 q.push(j);
             }
         }
     }
-    return ans.length == numCourses ? ans : [];
+    return ans.length === numCourses ? ans : [];
 }
 ```
 
 ```cs
 public class Solution {
     public int[] FindOrder(int numCourses, int[][] prerequisites) {
-        var g = new List<int>[numCourses];
-        for (int i = 0; i < numCourses; ++i)
-        {
+        List<int>[] g = new List<int>[numCourses];
+        for (int i = 0; i < numCourses; i++) {
             g[i] = new List<int>();
         }
-        var indeg = new int[numCourses];
-        foreach (var p in prerequisites)
-        {
+        int[] indeg = new int[numCourses];
+        foreach (var p in prerequisites) {
             int a = p[0], b = p[1];
             g[b].Add(a);
             ++indeg[a];
         }
-        var q = new Queue<int>();
-        for (int i = 0; i < numCourses; ++i)
-        {
-            if (indeg[i] == 0) q.Enqueue(i);
+        Queue<int> q = new Queue<int>();
+        for (int i = 0; i < numCourses; ++i) {
+            if (indeg[i] == 0) {
+                q.Enqueue(i);
+            }
         }
-        var ans = new int[numCourses];
-        var cnt = 0;
-        while (q.Count > 0)
-        {
+        int[] ans = new int[numCourses];
+        int cnt = 0;
+        while (q.Count > 0) {
             int i = q.Dequeue();
             ans[cnt++] = i;
-            foreach (int j in g[i])
-            {
-                if (--indeg[j] == 0) q.Enqueue(j);
+            foreach (int j in g[i]) {
+                if (--indeg[j] == 0) {
+                    q.Enqueue(j);
+                }
             }
         }
         return cnt == numCourses ? ans : new int[0];

lcof2/剑指 Offer II 113. 课程顺序/Solution.cpp

@@ -1,23 +1,29 @@
 class Solution {
 public:
     vector<int> findOrder(int numCourses, vector<vector<int>>& prerequisites) {
-        vector<vector<int>> g(numCourses);
+        vector<int> g[numCourses];
         vector<int> indeg(numCourses);
         for (auto& p : prerequisites) {
             int a = p[0], b = p[1];
             g[b].push_back(a);
             ++indeg[a];
         }
         queue<int> q;
-        for (int i = 0; i < numCourses; ++i)
-            if (indeg[i] == 0) q.push(i);
+        for (int i = 0; i < numCourses; ++i) {
+            if (indeg[i] == 0) {
+                q.push(i);
+            }
+        }
         vector<int> ans;
-        while (!q.empty()) {
+        while (q.size()) {
             int i = q.front();
             q.pop();
             ans.push_back(i);
-            for (int j : g[i])
-                if (--indeg[j] == 0) q.push(j);
+            for (int j : g[i]) {
+                if (--indeg[j] == 0) {
+                    q.push(j);
+                }
+            }
         }
         return ans.size() == numCourses ? ans : vector<int>();
     }

lcof2/剑指 Offer II 113. 课程顺序/Solution.cs

@@ -1,33 +1,32 @@
 public class Solution {
     public int[] FindOrder(int numCourses, int[][] prerequisites) {
-        var g = new List<int>[numCourses];
-        for (int i = 0; i < numCourses; ++i)
-        {
+        List<int>[] g = new List<int>[numCourses];
+        for (int i = 0; i < numCourses; i++) {
             g[i] = new List<int>();
         }
-        var indeg = new int[numCourses];
-        foreach (var p in prerequisites)
-        {
+        int[] indeg = new int[numCourses];
+        foreach (var p in prerequisites) {
             int a = p[0], b = p[1];
             g[b].Add(a);
             ++indeg[a];
         }
-        var q = new Queue<int>();
-        for (int i = 0; i < numCourses; ++i)
-        {
-            if (indeg[i] == 0) q.Enqueue(i);
+        Queue<int> q = new Queue<int>();
+        for (int i = 0; i < numCourses; ++i) {
+            if (indeg[i] == 0) {
+                q.Enqueue(i);
+            }
         }
-        var ans = new int[numCourses];
-        var cnt = 0;
-        while (q.Count > 0)
-        {
+        int[] ans = new int[numCourses];
+        int cnt = 0;
+        while (q.Count > 0) {
             int i = q.Dequeue();
             ans[cnt++] = i;
-            foreach (int j in g[i])
-            {
-                if (--indeg[j] == 0) q.Enqueue(j);
+            foreach (int j in g[i]) {
+                if (--indeg[j] == 0) {
+                    q.Enqueue(j);
+                }
             }
         }
         return cnt == numCourses ? ans : new int[0];
     }
-}
+}

lcof2/剑指 Offer II 113. 课程顺序/Solution.py

@@ -1,11 +1,11 @@
 class Solution:
     def findOrder(self, numCourses: int, prerequisites: List[List[int]]) -> List[int]:
-        g = defaultdict(list)
+        g = [[] for _ in range(numCourses)]
         indeg = [0] * numCourses
         for a, b in prerequisites:
             g[b].append(a)
             indeg[a] += 1
-        q = deque([i for i, v in enumerate(indeg) if v == 0])
+        q = deque(i for i, v in enumerate(indeg) if v == 0)
         ans = []
         while q:
             i = q.popleft()

lcof2/剑指 Offer II 113. 课程顺序/Solution.ts

@@ -1,25 +1,20 @@
 function findOrder(numCourses: number, prerequisites: number[][]): number[] {
-    let g = Array.from({ length: numCourses }, () => []);
-    let indeg = new Array(numCourses).fill(0);
-    for (let [a, b] of prerequisites) {
+    const g: number[][] = Array.from({ length: numCourses }, () => []);
+    const indeg: number[] = Array(numCourses).fill(0);
+    for (const [a, b] of prerequisites) {
         g[b].push(a);
         ++indeg[a];
     }
-    let q = [];
-    for (let i = 0; i < numCourses; ++i) {
-        if (!indeg[i]) {
-            q.push(i);
-        }
-    }
-    let ans = [];
+    const q: number[] = indeg.map((v, i) => (v === 0 ? i : -1)).filter(v => v !== -1);
+    const ans: number[] = [];
     while (q.length) {
-        const i = q.shift();
+        const i = q.pop()!;
         ans.push(i);
-        for (let j of g[i]) {
-            if (--indeg[j] == 0) {
+        for (const j of g[i]) {
+            if (--indeg[j] === 0) {
                 q.push(j);
             }
         }
     }
-    return ans.length == numCourses ? ans : [];
+    return ans.length === numCourses ? ans : [];
 }

requirements.txt

@@ -1,2 +1,2 @@
-black==24.2.0
+black==24.3.0
 Requests==2.31.0

solution/3000-3099/3060.User Activities within Time Bounds/README.md

@@ -1,4 +1,4 @@
-# [3060. User Activities within Time Bounds](https://leetcode.cn/problems/user-activities-within-time-bounds)
+# [3060. 时间范围内的用户活动](https://leetcode.cn/problems/user-activities-within-time-bounds)
 
 [English Version](/solution/3000-3099/3060.User%20Activities%20within%20Time%20Bounds/README_EN.md)
 
@@ -8,7 +8,7 @@
 
 <!-- 这里写题目描述 -->
 
-<p>Table: <code>Sessions</code></p>
+<p>表：<code>Sessions</code></p>
 
 <pre>
 +---------------+----------+
@@ -20,23 +20,24 @@
 | session_id    | int      |
 | session_type  | enum     |
 +---------------+----------+
-session_id is column of unique values for this table.
-session_type is an ENUM (category) type of (Viewer, Streamer).
-This table contains user id, session start, session end, session id and session type.
+session_id 是这张表中有不同值的列。
+session_type 是 (Viewer, Streamer) 的一个 ENUM (category) 类型。
+这张表包含 user id, session start, session end, session id 和 session 类型。
 </pre>
 
-<p>Write a solution to find the the <strong>users</strong> who have had <strong>at least one</strong> <strong>consecutive session</strong> of the <strong>same</strong> type (either &#39;<strong>Viewer</strong>&#39; or &#39;<strong>Streamer</strong>&#39;) with a <strong>maximum</strong> gap of <code>12</code> hours <strong>between</strong> sessions.</p>
+<p>编写一个解决方案，以查找 <strong>至少有一个相同</strong> 类型的 <strong>连续会话</strong>（无论是“<strong>Viewer</strong>”还是“<strong>Streamer</strong>”）的 <strong>用户</strong>，会话 <strong>之间</strong> 的 <strong>最大</strong> 间隔为 <code>12</code> 小时。</p>
 
-<p>Return <em>the result table ordered by </em><code>user_id</code><em> in <b>ascending</b> order.</em></p>
+<p>返回结果表，以<em>&nbsp;</em><code>user_id</code><em>&nbsp;<strong>升序</strong> 排序。</em></p>
 
-<p>The result format is in the following example.</p>
+<p>结果格式如下所述。</p>
 
 <p>&nbsp;</p>
-<p><strong class="example">Example:</strong></p>
+
+<p><strong class="example">示例：</strong></p>
 
 <pre>
-<strong>Input:</strong> 
-Sessions table:
+<strong>输入:</strong> 
+Sessions 表:
 +---------+---------------------+---------------------+------------+--------------+
 | user_id | session_start       | session_end         | session_id | session_type | 
 +---------+---------------------+---------------------+------------+--------------+
@@ -52,18 +53,18 @@ Sessions table:
 | 103     | 2023-11-02 20:00:00 | 2023-11-02 23:00:00 | 10         | Viewer       | 
 | 103     | 2023-11-03 09:00:00 | 2023-11-03 10:00:00 | 11         | Viewer       | 
 +---------+---------------------+---------------------+------------+--------------+
-<strong>Output:</strong> 
+<strong>输出:</strong> 
 +---------+
 | user_id |
 +---------+
 | 102     |
 | 103     |
 +---------+
-<strong>Explanation:</strong>
-- User ID 101 will not be included in the final output as they do not have any consecutive sessions of the same session type.
-- User ID 102 will be included in the final output as they had two viewer sessions with session IDs 3 and 4, respectively, and the time gap between them was less than 12 hours.
-- User ID 103 participated in two viewer sessions with a gap of less than 12 hours between them, identified by session IDs 10 and 11. Therefore, user 103 will be included in the final output.
-Output table is ordered by user_id in increasing order.
+<strong>解释:</strong>
+- 用户 ID 101 将不会包含在最终输出中，因为他们没有相同会话类型的连续回话。
+- 用户 ID 102 将会包含在最终输出中，因为他们分别有两个 session ID 为 3 和 4 的 viewer 会话，并且时间间隔在 12 小时内。
+- 用户 ID 103 参与了两次 viewer 会话，间隔不到 12 小时，会话 ID 为 10 和 11。因此，用户 103 将会被包含在最终输出中。
+输出表根据 user_id 升序排列。
 </pre>
 
 ## 解法

solution/3000-3099/3063.Linked List Frequency/README.md

@@ -8,7 +8,7 @@
 
 <!-- 这里写题目描述 -->
 
-<p>给定包含 <code>k</code> 个&nbsp;<strong>不同&nbsp;</strong>元素的链表的&nbsp;<code>head</code>&nbsp;节点，创建一个长度为&nbsp;<code>k</code>&nbsp;的链表，包含给定链表中每个 <strong>不同元素</strong> 以 <strong>任何顺序</strong> 出现的 <span data-keyword="frequency-linkedlist">频率</span>&nbsp;。返回这个链表的头节点。</p>
+<p>给定包含 <code>k</code> 个&nbsp;<strong>不同&nbsp;</strong>元素的链表的&nbsp;<code>head</code>&nbsp;节点，创建一个长度为&nbsp;<code>k</code>&nbsp;的链表，以 <strong>任何顺序</strong> 返回链表中所有 <strong>不同元素</strong> 出现的 <strong>频率</strong>。返回这个链表的头节点。</p>
 
 <p>&nbsp;</p>