feat: add solutions to leetcode problem: No.0356. Line Reflection

* add leetcode solutions: No.0356. Line Reflection
* fix images path

Author: yanglbme

solution/0200-0299/0277.Find the Celebrity/README.md

@@ -18,7 +18,7 @@
 
 <p><strong>示例 1:</strong></p>
 
-<p><img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0200-0299/0277.Find%20the%20Celebrity/images/277_example_1_bold.PNG" style="height: 181px; width: 186px;" /></p>
+<p><img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0200-0299/0277.Find%20the%20Celebrity/images/277_example_1_bold.png" style="height: 181px; width: 186px;" /></p>
 
 <pre>
 <strong>输入: </strong>graph = [
@@ -32,7 +32,7 @@
 
 <p><strong>示例 2:</strong></p>
 
-<p><img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0200-0299/0277.Find%20the%20Celebrity/images/277_example_2.PNG" style="height: 192px; width: 193px;" /></p>
+<p><img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0200-0299/0277.Find%20the%20Celebrity/images/277_example_2.png" style="height: 192px; width: 193px;" /></p>
 
 <pre>
 <strong>输入: </strong>graph = [

solution/0200-0299/0277.Find the Celebrity/README_EN.md

@@ -12,15 +12,15 @@
 
 <p>&nbsp;</p>
 <p><strong>Example 1:</strong></p>
-<img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0200-0299/0277.Find%20the%20Celebrity/images/277_example_1_bold.PNG" style="width: 186px; height: 181px;" />
+<img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0200-0299/0277.Find%20the%20Celebrity/images/277_example_1_bold.png" style="width: 186px; height: 181px;" />
 <pre>
 <strong>Input:</strong> graph = [[1,1,0],[0,1,0],[1,1,1]]
 <strong>Output:</strong> 1
 <strong>Explanation:</strong> There are three persons labeled with 0, 1 and 2. graph[i][j] = 1 means person i knows person j, otherwise graph[i][j] = 0 means person i does not know person j. The celebrity is the person labeled as 1 because both 0 and 2 know him but 1 does not know anybody.
 </pre>
 
 <p><strong>Example 2:</strong></p>
-<img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0200-0299/0277.Find%20the%20Celebrity/images/277_example_2.PNG" style="width: 193px; height: 192px;" />
+<img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0200-0299/0277.Find%20the%20Celebrity/images/277_example_2.png" style="width: 193px; height: 192px;" />
 <pre>
 <strong>Input:</strong> graph = [[1,0,1],[1,1,0],[0,1,1]]
 <strong>Output:</strong> -1

solution/0200-0299/0285.Inorder Successor in BST/README.md

@@ -14,7 +14,7 @@
 
 <p><strong>示例 1：</strong></p>
 
-<p><img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0200-0299/0285.Inorder%20Successor%20in%20BST/images/285_example_1.PNG" style="height: 117px; width: 122px;" /></p>
+<p><img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0200-0299/0285.Inorder%20Successor%20in%20BST/images/285_example_1.png" style="height: 117px; width: 122px;" /></p>
 
 <pre>
 <strong>输入：</strong>root = [2,1,3], p = 1
@@ -24,7 +24,7 @@
 
 <p><strong>示例 2：</strong></p>
 
-<p><img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0200-0299/0285.Inorder%20Successor%20in%20BST/images/285_example_2.PNG" style="height: 229px; width: 246px;" /></p>
+<p><img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0200-0299/0285.Inorder%20Successor%20in%20BST/images/285_example_2.png" style="height: 229px; width: 246px;" /></p>
 
 <pre>
 <strong>输入：</strong>root = [5,3,6,2,4,null,null,1], p = 6

solution/0200-0299/0285.Inorder Successor in BST/README_EN.md

@@ -10,15 +10,15 @@
 
 <p>&nbsp;</p>
 <p><strong>Example 1:</strong></p>
-<img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0200-0299/0285.Inorder%20Successor%20in%20BST/images/285_example_1.PNG" style="width: 122px; height: 117px;" />
+<img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0200-0299/0285.Inorder%20Successor%20in%20BST/images/285_example_1.png" style="width: 122px; height: 117px;" />
 <pre>
 <strong>Input:</strong> root = [2,1,3], p = 1
 <strong>Output:</strong> 2
 <strong>Explanation:</strong> 1&#39;s in-order successor node is 2. Note that both p and the return value is of TreeNode type.
 </pre>
 
 <p><strong>Example 2:</strong></p>
-<img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0200-0299/0285.Inorder%20Successor%20in%20BST/images/285_example_2.PNG" style="width: 246px; height: 229px;" />
+<img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0200-0299/0285.Inorder%20Successor%20in%20BST/images/285_example_2.png" style="width: 246px; height: 229px;" />
 <pre>
 <strong>Input:</strong> root = [5,3,6,2,4,null,null,1], p = 6
 <strong>Output:</strong> null

solution/0300-0399/0356.Line Reflection/README.md

@@ -17,15 +17,15 @@
 <p> </p>
 
 <p><strong>示例 1：</strong></p>
-<img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0300-0399/0356.Line%20Reflection/images/356_example_1.PNG" style="width: 389px; height: 340px;" />
+<img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0300-0399/0356.Line%20Reflection/images/356_example_1.png" style="width: 389px; height: 340px;" />
 <pre>
 <strong>输入：</strong>points = [[1,1],[-1,1]]
 <strong>输出：</strong>true
 <strong>解释：</strong>可以找出 x = 0 这条线。
 </pre>
 
 <p><strong>示例 2：</strong></p>
-<img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0300-0399/0356.Line%20Reflection/images/356_example_2.PNG" style="width: 300px; height: 294px;" />
+<img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0300-0399/0356.Line%20Reflection/images/356_example_2.png" style="width: 300px; height: 294px;" />
 <pre>
 <strong>输入：</strong>points = [[1,1],[-1,-1]]
 <strong>输出：</strong>false
@@ -41,27 +41,59 @@
 	<li><code>-10^8 <= points[i][j] <= 10^8</code></li>
 </ul>
 
-
 ## 解法
 
 <!-- 这里可写通用的实现逻辑 -->
 
+先找出所有点中的最小、最大的 x 坐标 `minX` 和 `maxX`。若存在满足条件的直线，则直线 `x = (minX + maxX) / 2`。(或者说：`s = minX + maxX`)
+
+遍历每个点 `point(x, y)`，若 (s - x, y) 不在点集里，说明不满足条件，直接返回 false。遍历结束返回 true。
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def isReflected(self, points: List[List[int]]) -> bool:
+        min_x, max_x = float('inf'), float('-inf')
+        point_set = set()
+        for point in points:
+            min_x = min(min_x, point[0])
+            max_x = max(max_x, point[0])
+            point_set.add((point[0], point[1]))
+        s = min_x + max_x
+        for point in points:
+            if (s - point[0], point[1]) not in point_set:
+                return False
+        return True
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
-
+class Solution {
+    public boolean isReflected(int[][] points) {
+        int minX = Integer.MAX_VALUE, maxX = Integer.MIN_VALUE;
+        Set<String> pointSet = new HashSet<>();
+        for (int[] point : points) {
+            minX = Math.min(minX, point[0]);
+            maxX = Math.max(maxX, point[0]);
+            pointSet.add(point[0] + "." + point[1]);
+        }
+        long s = minX + maxX;
+        for (int[] point : points) {
+            if (!pointSet.contains((s - point[0]) + "." + point[1])) {
+                return false;
+            }
+        }
+        return true;
+    }
+}
 ```
 
 ### **...**

solution/0300-0399/0356.Line Reflection/README_EN.md

@@ -13,15 +13,15 @@ Could you do better than O(<i>n</i><sup>2</sup>) ?</p>
 
 <p>&nbsp;</p>
 <p><strong>Example 1:</strong></p>
-<img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0300-0399/0356.Line%20Reflection/images/356_example_1.PNG" style="width: 389px; height: 340px;" />
+<img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0300-0399/0356.Line%20Reflection/images/356_example_1.png" style="width: 389px; height: 340px;" />
 <pre>
 <strong>Input:</strong> points = [[1,1],[-1,1]]
 <strong>Output:</strong> true
 <strong>Explanation:</strong> We can choose the line x = 0.
 </pre>
 
 <p><strong>Example 2:</strong></p>
-<img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0300-0399/0356.Line%20Reflection/images/356_example_2.PNG" style="width: 300px; height: 294px;" />
+<img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0300-0399/0356.Line%20Reflection/images/356_example_2.png" style="width: 300px; height: 294px;" />
 <pre>
 <strong>Input:</strong> points = [[1,1],[-1,-1]]
 <strong>Output:</strong> false
@@ -45,13 +45,42 @@ Could you do better than O(<i>n</i><sup>2</sup>) ?</p>
 ### **Python3**
 
 ```python
-
+class Solution:
+    def isReflected(self, points: List[List[int]]) -> bool:
+        min_x, max_x = float('inf'), float('-inf')
+        point_set = set()
+        for point in points:
+            min_x = min(min_x, point[0])
+            max_x = max(max_x, point[0])
+            point_set.add((point[0], point[1]))
+        s = min_x + max_x
+        for point in points:
+            if (s - point[0], point[1]) not in point_set:
+                return False
+        return True
 ```
 
 ### **Java**
 
 ```java
-
+class Solution {
+    public boolean isReflected(int[][] points) {
+        int minX = Integer.MAX_VALUE, maxX = Integer.MIN_VALUE;
+        Set<String> pointSet = new HashSet<>();
+        for (int[] point : points) {
+            minX = Math.min(minX, point[0]);
+            maxX = Math.max(maxX, point[0]);
+            pointSet.add(point[0] + "." + point[1]);
+        }
+        long s = minX + maxX;
+        for (int[] point : points) {
+            if (!pointSet.contains((s - point[0]) + "." + point[1])) {
+                return false;
+            }
+        }
+        return true;
+    }
+}
 ```
 
 ### **...**

solution/0300-0399/0356.Line Reflection/Solution.java

@@ -0,0 +1,18 @@
+class Solution {
+    public boolean isReflected(int[][] points) {
+        int minX = Integer.MAX_VALUE, maxX = Integer.MIN_VALUE;
+        Set<String> pointSet = new HashSet<>();
+        for (int[] point : points) {
+            minX = Math.min(minX, point[0]);
+            maxX = Math.max(maxX, point[0]);
+            pointSet.add(point[0] + "." + point[1]);
+        }
+        long s = minX + maxX;
+        for (int[] point : points) {
+            if (!pointSet.contains((s - point[0]) + "." + point[1])) {
+                return false;
+            }
+        }
+        return true;
+    }
+}

solution/0300-0399/0356.Line Reflection/Solution.py

@@ -0,0 +1,13 @@
+class Solution:
+    def isReflected(self, points: List[List[int]]) -> bool:
+        min_x, max_x = float('inf'), float('-inf')
+        point_set = set()
+        for point in points:
+            min_x = min(min_x, point[0])
+            max_x = max(max_x, point[0])
+            point_set.add((point[0], point[1]))
+        s = min_x + max_x
+        for point in points:
+            if (s - point[0], point[1]) not in point_set:
+                return False
+        return True

solution/0500-0599/0510.Inorder Successor in BST II/README.md

@@ -24,7 +24,7 @@ class Node {
 
 <p><strong>示例 1：</strong></p>
 
-<p><img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0500-0599/0510.Inorder%20Successor%20in%20BST%20II/images/285_example_1.PNG" style="height: 117px; width: 122px;" /></p>
+<p><img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0500-0599/0510.Inorder%20Successor%20in%20BST%20II/images/285_example_1.png" style="height: 117px; width: 122px;" /></p>
 
 <pre>
 <strong>输入：</strong>tree = [2,1,3], node = 1
@@ -34,7 +34,7 @@ class Node {
 
 <p><strong>示例 2：</strong></p>
 
-<p><img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0500-0599/0510.Inorder%20Successor%20in%20BST%20II/images/285_example_2.PNG" style="height: 229px; width: 246px;" /></p>
+<p><img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0500-0599/0510.Inorder%20Successor%20in%20BST%20II/images/285_example_2.png" style="height: 229px; width: 246px;" /></p>
 
 <pre>
 <strong>输入：</strong>tree = [5,3,6,2,4,null,null,1], node = 6
@@ -44,7 +44,7 @@ class Node {
 
 <p><strong>示例 3：</strong></p>
 
-<p><img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0500-0599/0510.Inorder%20Successor%20in%20BST%20II/images/285_example_34.PNG" style="height: 335px; width: 438px;" /></p>
+<p><img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0500-0599/0510.Inorder%20Successor%20in%20BST%20II/images/285_example_34.png" style="height: 335px; width: 438px;" /></p>
 
 <pre>
 <strong>输入：</strong>tree = [15,6,18,3,7,17,20,2,4,null,13,null,null,null,null,null,null,null,null,9], node = 15
@@ -53,7 +53,7 @@ class Node {
 
 <p><strong>示例 4：</strong></p>
 
-<p><img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0500-0599/0510.Inorder%20Successor%20in%20BST%20II/images/285_example_34.PNG" style="height: 335px; width: 438px;" /></p>
+<p><img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0500-0599/0510.Inorder%20Successor%20in%20BST%20II/images/285_example_34.png" style="height: 335px; width: 438px;" /></p>
 
 <pre>
 <strong>输入：</strong>tree = [15,6,18,3,7,17,20,2,4,null,13,null,null,null,null,null,null,null,null,9], node = 13

solution/0500-0599/0510.Inorder Successor in BST II/README_EN.md

@@ -21,30 +21,30 @@ class Node {
 
 <p>&nbsp;</p>
 <p><strong>Example 1:</strong></p>
-<img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0500-0599/0510.Inorder%20Successor%20in%20BST%20II/images/285_example_1.PNG" style="width: 122px; height: 117px;" />
+<img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0500-0599/0510.Inorder%20Successor%20in%20BST%20II/images/285_example_1.png" style="width: 122px; height: 117px;" />
 <pre>
 <strong>Input:</strong> tree = [2,1,3], node = 1
 <strong>Output:</strong> 2
 <strong>Explanation:</strong> 1&#39;s in-order successor node is 2. Note that both the node and the return value is of Node type.
 </pre>
 
 <p><strong>Example 2:</strong></p>
-<img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0500-0599/0510.Inorder%20Successor%20in%20BST%20II/images/285_example_2.PNG" style="width: 246px; height: 229px;" />
+<img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0500-0599/0510.Inorder%20Successor%20in%20BST%20II/images/285_example_2.png" style="width: 246px; height: 229px;" />
 <pre>
 <strong>Input:</strong> tree = [5,3,6,2,4,null,null,1], node = 6
 <strong>Output:</strong> null
 <strong>Explanation:</strong> There is no in-order successor of the current node, so the answer is null.
 </pre>
 
 <p><strong>Example 3:</strong></p>
-<img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0500-0599/0510.Inorder%20Successor%20in%20BST%20II/images/285_example_34.PNG" style="width: 438px; height: 335px;" />
+<img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0500-0599/0510.Inorder%20Successor%20in%20BST%20II/images/285_example_34.png" style="width: 438px; height: 335px;" />
 <pre>
 <strong>Input:</strong> tree = [15,6,18,3,7,17,20,2,4,null,13,null,null,null,null,null,null,null,null,9], node = 15
 <strong>Output:</strong> 17
 </pre>
 
 <p><strong>Example 4:</strong></p>
-<img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0500-0599/0510.Inorder%20Successor%20in%20BST%20II/images/285_example_34.PNG" style="width: 438px; height: 335px;" />
+<img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0500-0599/0510.Inorder%20Successor%20in%20BST%20II/images/285_example_34.png" style="width: 438px; height: 335px;" />
 <pre>
 <strong>Input:</strong> tree = [15,6,18,3,7,17,20,2,4,null,13,null,null,null,null,null,null,null,null,9], node = 13
 <strong>Output:</strong> 15