[pull] main from doocs:main

README_EN.md

@@ -14,7 +14,9 @@
 
 ## Introduction
 
-Complete solutions to LeetCode, LCOF and LCCI problems, updated daily. Please give me a [star](https://github.com/doocs/leetcode) 🌟 if you like it.
+The Doocs LeetCode repository is a comprehensive collection of solutions to LeetCode questions in multiple programming languages. The repository contains solutions to LeetCode, LCOF, LCCI questions, and more in multiple programming languages.
+
+The repository is maintained by the Doocs community, and please give us a [star](https://github.com/doocs/leetcode) 🌟 if you like it.
 
 [中文文档](/README.md)
 

solution/0000-0099/0014.Longest Common Prefix/README.md

@@ -229,6 +229,27 @@ impl Solution {
 }
 ```
 
+### **PHP**
+
+```php
+class Solution {
+    /**
+     * @param String[] $strs
+     * @return String
+     */
+    function longestCommonPrefix($strs) {
+        $rs = "";
+        for ($i = 0; $i < strlen($strs[0]); $i++) {
+            for ($j = 1; $j < count($strs); $j++) {
+                if ($strs[0][$i] != $strs[$j][$i]) return $rs;
+            }
+            $rs = $rs.$strs[0][$i];
+        }
+        return $rs;
+    }
+}
+```
+
 ### **...**
 
 ```

solution/0000-0099/0014.Longest Common Prefix/README_EN.md

@@ -220,6 +220,27 @@ impl Solution {
 }
 ```
 
+### **PHP**
+
+```php
+class Solution {
+    /**
+     * @param String[] $strs
+     * @return String
+     */
+    function longestCommonPrefix($strs) {
+        $rs = "";
+        for ($i = 0; $i < strlen($strs[0]); $i++) {
+            for ($j = 1; $j < count($strs); $j++) {
+                if ($strs[0][$i] != $strs[$j][$i]) return $rs;
+            }
+            $rs = $rs.$strs[0][$i];
+        }
+        return $rs;
+    }
+}
+```
+
 ### **...**
 
 ```

solution/0000-0099/0014.Longest Common Prefix/Solution.php

@@ -0,0 +1,16 @@
+class Solution {
+    /**
+     * @param String[] $strs
+     * @return String
+     */
+    function longestCommonPrefix($strs) {
+        $rs = "";
+        for ($i = 0; $i < strlen($strs[0]); $i++) {
+            for ($j = 1; $j < count($strs); $j++) {
+                if ($strs[0][$i] != $strs[$j][$i]) return $rs;
+            }
+            $rs = $rs.$strs[0][$i];
+        }
+        return $rs;
+    }
+}

solution/0200-0299/0207.Course Schedule/README.md

@@ -50,7 +50,13 @@
 
 **方法一：拓扑排序**
 
-BFS 实现。
+对于本题，我们可以将课程看作图中的节点，先修课程看作图中的边，那么我们可以将本题转化为判断有向图中是否存在环。
+
+具体地，我们可以使用拓扑排序的思想，对于每个入度为 $0$ 的节点，我们将其出度的节点的入度减 $1$，直到所有节点都被遍历到。
+
+如果所有节点都被遍历到，说明图中不存在环，那么我们就可以完成所有课程的学习；否则，我们就无法完成所有课程的学习。
+
+时间复杂度 $O(n + m)$，空间复杂度 $O(n + m)$。其中 $n$ 和 $m$ 分别为课程数和先修课程数。
 
 <!-- tabs:start -->
 
@@ -67,7 +73,7 @@ class Solution:
             g[b].append(a)
             indeg[a] += 1
         cnt = 0
-        q = deque([i for i, v in enumerate(indeg) if v == 0])
+        q = deque(i for i, x in enumerate(indeg) if x == 0)
         while q:
             i = q.popleft()
             cnt += 1
@@ -114,36 +120,6 @@ class Solution {
 }
 ```
 
-### **TypeScrpt**
-
-```ts
-function canFinish(numCourses: number, prerequisites: number[][]): boolean {
-    let g = Array.from({ length: numCourses }, () => []);
-    let indeg = new Array(numCourses).fill(0);
-    for (let [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 cnt = 0;
-    while (q.length) {
-        const i = q.shift();
-        ++cnt;
-        for (let j of g[i]) {
-            if (--indeg[j] == 0) {
-                q.push(j);
-            }
-        }
-    }
-    return cnt == numCourses;
-}
-```
-
 ### **C++**
 
 ```cpp
@@ -158,15 +134,21 @@ public:
             ++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);
+            }
+        }
         int cnt = 0;
         while (!q.empty()) {
             int i = q.front();
             q.pop();
             ++cnt;
-            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 cnt == numCourses;
     }
@@ -185,8 +167,8 @@ func canFinish(numCourses int, prerequisites [][]int) bool {
 		indeg[a]++
 	}
 	q := []int{}
-	for i, v := range indeg {
-		if v == 0 {
+	for i, x := range indeg {
+		if x == 0 {
 			q = append(q, i)
 		}
 	}
@@ -206,36 +188,65 @@ func canFinish(numCourses int, prerequisites [][]int) bool {
 }
 ```
 
+### **TypeScript**
+
+```ts
+function canFinish(numCourses: number, prerequisites: number[][]): boolean {
+    const g: number[][] = new Array(numCourses).fill(0).map(() => []);
+    const indeg: number[] = new Array(numCourses).fill(0);
+    for (const [a, b] of prerequisites) {
+        g[b].push(a);
+        indeg[a]++;
+    }
+    const q: number[] = [];
+    for (let i = 0; i < numCourses; ++i) {
+        if (indeg[i] == 0) {
+            q.push(i);
+        }
+    }
+    let cnt = 0;
+    while (q.length) {
+        const i = q.shift()!;
+        cnt++;
+        for (const j of g[i]) {
+            if (--indeg[j] == 0) {
+                q.push(j);
+            }
+        }
+    }
+    return cnt == numCourses;
+}
+```
+
 ### **C#**
 
 ```cs
 public class Solution {
     public bool CanFinish(int numCourses, int[][] prerequisites) {
         var g = new List<int>[numCourses];
-        for (int i = 0; i < numCourses; ++i)
-        {
+        for (int i = 0; i < numCourses; ++i) {
             g[i] = new List<int>();
         }
         var indeg = new int[numCourses];
-        foreach (var p in prerequisites)
-        {
+        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);
+        for (int i = 0; i < numCourses; ++i) {
+            if (indeg[i] == 0) {
+                q.Enqueue(i);
+            }
         }
         var cnt = 0;
-        while (q.Count > 0)
-        {
+        while (q.Count > 0) {
             int i = q.Dequeue();
             ++cnt;
-            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;