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feat: add solutions to lc problems: No.1886,1887,1888,1889
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‎solution/0200-0299/0237.Delete Node in a Linked List/README.md

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@@ -128,4 +128,24 @@ func deleteNode(node *ListNode) {
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}
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```
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### **C++**
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```cpp
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/**
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* Definition for singly-linked list.
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* struct ListNode {
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* int val;
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* ListNode *next;
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* ListNode(int x) : val(x), next(NULL) {}
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* };
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*/
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class Solution {
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public:
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void deleteNode(ListNode* node) {
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node->val = node->next->val;
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node->next = node->next->next;
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}
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};
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```
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‎solution/0200-0299/0237.Delete Node in a Linked List/README_EN.md

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}
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```
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### **C++**
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```cpp
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/**
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* Definition for singly-linked list.
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* struct ListNode {
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* int val;
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* ListNode *next;
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* ListNode(int x) : val(x), next(NULL) {}
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* };
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*/
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class Solution {
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public:
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void deleteNode(ListNode* node) {
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node->val = node->next->val;
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node->next = node->next->next;
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}
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};
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```
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<!-- tabs:end -->

‎solution/0200-0299/0237.Delete Node in a Linked List/Solution.c

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/**
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* Definition for singly-linked list.
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* struct ListNode {
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* int val;
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* struct ListNode *next;
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* };
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*/
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void deleteNode(struct ListNode* node) {
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node->val = node->next->val;
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node->next = node->next->next;
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}

‎solution/0200-0299/0237.Delete Node in a Linked List/Solution.cpp

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/**
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* Definition for singly-linked list.
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* struct ListNode {
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* int val;
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* ListNode *next;
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* ListNode(int x) : val(x), next(NULL) {}
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* };
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*/
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class Solution {
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public:
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void deleteNode(ListNode* node) {
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node->val = node->next->val;
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node->next = node->next->next;
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}
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};

‎solution/1800-1899/1884.Egg Drop With 2 Eggs and N Floors/README.md

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# [1884. ](https://leetcode-cn.com/problems/egg-drop-with-2-eggs-and-n-floors)
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# [1884. 鸡蛋掉落-两枚鸡蛋](https://leetcode-cn.com/problems/egg-drop-with-2-eggs-and-n-floors)
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[English Version](/solution/1800-1899/1884.Egg%20Drop%20With%202%20Eggs%20and%20N%20Floors/README_EN.md)
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## 题目描述
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<!-- 这里写题目描述 -->
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None
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<p>给你 <strong>2 枚相同 </strong>的鸡蛋，和一栋从第 <code>1</code> 层到第 <code>n</code> 层共有 <code>n</code> 层楼的建筑。</p>
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<p>已知存在楼层 <code>f</code> ，满足 <code>0 <= f <= n</code> ，任何从 <strong>高于 </strong><code>f</code> 的楼层落下的鸡蛋都<strong> 会碎 </strong>，从 <strong><code>f</code> 楼层或比它低 </strong>的楼层落下的鸡蛋都 <strong>不会碎 </strong>。</p>
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<p>每次操作，你可以取一枚<strong> 没有碎</strong> 的鸡蛋并把它从任一楼层 <code>x</code> 扔下（满足 <code>1 <= x <= n</code>）。如果鸡蛋碎了，你就不能再次使用它。如果某枚鸡蛋扔下后没有摔碎，则可以在之后的操作中<strong> 重复使用 </strong>这枚鸡蛋。</p>
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<p>请你计算并返回要确定 <code>f</code> <strong>确切的值 </strong>的 <strong>最小操作次数</strong> 是多少？</p>
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<p> </p>
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<p><strong>示例 1：</strong></p>
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<pre>
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<strong>输入：</strong>n = 2
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<strong>输出：</strong>2
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<strong>解释：</strong>我们可以将第一枚鸡蛋从 1 楼扔下，然后将第二枚从 2 楼扔下。
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如果第一枚鸡蛋碎了，可知 f = 0；
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如果第二没鸡蛋碎了，但第一枚没碎，可知 f = 1；
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否则，当两个鸡蛋都没碎时，可知 f = 2。
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</pre>
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<p><strong>示例 2：</strong></p>
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<pre>
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<strong>输入：</strong>n = 100
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<strong>输出：</strong>14
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</pre>
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<p> </p>
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<p><strong>提示：</strong></p>
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<ul>
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<li><code>1 <= n <= 1000</code></li>
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</ul>
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## 解法
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‎solution/1800-1899/1886.Determine Whether Matrix Can Be Obtained By Rotation/README.md

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# [1886. 判断矩阵经轮转后是否一致](https://leetcode-cn.com/problems/determine-whether-matrix-can-be-obtained-by-rotation)
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[English Version](/solution/1800-1899/1886.Determine%20Whether%20Matrix%20Can%20Be%20Obtained%20By%20Rotation/README_EN.md)
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## 题目描述
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<!-- 这里写题目描述 -->
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<p>给你两个大小为 <code>n x n</code> 的二进制矩阵 <code>mat</code> 和 <code>target</code> 。现<strong> 以 90 度顺时针轮转 </strong>矩阵 <code>mat</code> 中的元素 <strong>若干次</strong> ，如果能够使 <code>mat</code> 与 <code>target</code> 一致，返回 <code>true</code> ；否则，返回<em> </em><code>false</code><em> 。</em></p>
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<p> </p>
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<p><strong>示例 1：</strong></p>
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<img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/1800-1899/1886.Determine%20Whether%20Matrix%20Can%20Be%20Obtained%20By%20Rotation/images/grid3.png" style="width: 301px; height: 121px;" />
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<pre>
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<strong>输入：</strong>mat = [[0,1],[1,0]], target = [[1,0],[0,1]]
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<strong>输出：</strong>true
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<strong>解释：</strong>顺时针轮转 90 度一次可以使 mat 和 target 一致。
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</pre>
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<p><strong>示例 2：</strong></p>
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<img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/1800-1899/1886.Determine%20Whether%20Matrix%20Can%20Be%20Obtained%20By%20Rotation/images/grid4.png" style="width: 301px; height: 121px;" />
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<pre>
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<strong>输入：</strong>mat = [[0,1],[1,1]], target = [[1,0],[0,1]]
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<strong>输出：</strong>false
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<strong>解释：</strong>无法通过轮转矩阵中的元素使 equal 与 target 一致。
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</pre>
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<p><strong>示例 3：</strong></p>
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<img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/1800-1899/1886.Determine%20Whether%20Matrix%20Can%20Be%20Obtained%20By%20Rotation/images/grid4-1.png" style="width: 661px; height: 184px;" />
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<pre>
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<strong>输入：</strong>mat = [[0,0,0],[0,1,0],[1,1,1]], target = [[1,1,1],[0,1,0],[0,0,0]]
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<strong>输出：</strong>true
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<strong>解释：</strong>顺时针轮转 90 度两次可以使 mat 和 target 一致。
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</pre>
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<p> </p>
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<p><strong>提示：</strong></p>
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<ul>
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<li><code>n == mat.length == target.length</code></li>
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<li><code>n == mat[i].length == target[i].length</code></li>
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<li><code>1 <= n <= 10</code></li>
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<li><code>mat[i][j]</code> 和 <code>target[i][j]</code> 不是 <code>0</code> 就是 <code>1</code></li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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旋转矩阵，判断矩阵是否一致。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def findRotation(self, mat: List[List[int]], target: List[List[int]]) -> bool:
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def rotate(matrix):
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n = len(matrix)
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for i in range(n // 2):
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for j in range(i, n - 1 - i):
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t = matrix[i][j]
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matrix[i][j] = matrix[n - j - 1][i]
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matrix[n - j - 1][i] = matrix[n - i - 1][n - j - 1]
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matrix[n - i - 1][n - j - 1] = matrix[j][n - i - 1]
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matrix[j][n - i - 1] = t
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for _ in range(4):
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if mat == target:
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return True
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rotate(mat)
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return False
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public boolean findRotation(int[][] mat, int[][] target) {
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int times = 4;
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while (times-- > 0) {
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if (equals(mat, target)) {
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return true;
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}
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rotate(mat);
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}
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return false;
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}
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private void rotate(int[][] matrix) {
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int n = matrix.length;
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for (int i = 0; i < n / 2; ++i) {
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for (int j = i; j < n - 1 - i; ++j) {
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int t = matrix[i][j];
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matrix[i][j] = matrix[n - j - 1][i];
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matrix[n - j - 1][i] = matrix[n - i - 1][n - j - 1];
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matrix[n - i - 1][n - j - 1] = matrix[j][n - i - 1];
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matrix[j][n - i - 1] = t;
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}
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}
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}
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private boolean equals(int[][] nums1, int[][] nums2) {
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int n = nums1.length;
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for (int i = 0; i < n; ++i) {
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for (int j = 0; j < n; ++j) {
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if (nums1[i][j] != nums2[i][j]) {
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return false;
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}
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}
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}
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return true;
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}
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}
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```
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### **...**
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```
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```
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<!-- tabs:end -->

‎solution/1800-1899/1886.Determine Whether Matrix Can Be Obtained By Rotation/README_EN.md

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# [1886. Determine Whether Matrix Can Be Obtained By Rotation](https://leetcode.com/problems/determine-whether-matrix-can-be-obtained-by-rotation)
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[中文文档](/solution/1800-1899/1886.Determine%20Whether%20Matrix%20Can%20Be%20Obtained%20By%20Rotation/README.md)
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## Description
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<p>Given two <code>n x n</code> binary matrices <code>mat</code> and <code>target</code>, return <code>true</code><em> if it is possible to make </em><code>mat</code><em> equal to </em><code>target</code><em> by <strong>rotating</strong> </em><code>mat</code><em> in <strong>90-degree increments</strong>, or </em><code>false</code><em> otherwise.</em></p>
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<p>&nbsp;</p>
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<p><strong>Example 1:</strong></p>
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<img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/1800-1899/1886.Determine%20Whether%20Matrix%20Can%20Be%20Obtained%20By%20Rotation/images/grid3.png" style="width: 301px; height: 121px;" />
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<pre>
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<strong>Input:</strong> mat = [[0,1],[1,0]], target = [[1,0],[0,1]]
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<strong>Output:</strong> true
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<strong>Explanation: </strong>We can rotate mat 90 degrees clockwise to make mat equal target.
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</pre>
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<p><strong>Example 2:</strong></p>
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<img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/1800-1899/1886.Determine%20Whether%20Matrix%20Can%20Be%20Obtained%20By%20Rotation/images/grid4.png" style="width: 301px; height: 121px;" />
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<pre>
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<strong>Input:</strong> mat = [[0,1],[1,1]], target = [[1,0],[0,1]]
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<strong>Output:</strong> false
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<strong>Explanation:</strong> It is impossible to make mat equal to target by rotating mat.
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</pre>
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<p><strong>Example 3:</strong></p>
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<img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/1800-1899/1886.Determine%20Whether%20Matrix%20Can%20Be%20Obtained%20By%20Rotation/images/grid4-1.png" style="width: 661px; height: 184px;" />
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<pre>
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<strong>Input:</strong> mat = [[0,0,0],[0,1,0],[1,1,1]], target = [[1,1,1],[0,1,0],[0,0,0]]
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<strong>Output:</strong> true
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<strong>Explanation: </strong>We can rotate mat 90 degrees clockwise two times to make mat equal target.
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>n == mat.length == target.length</code></li>
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<li><code>n == mat[i].length == target[i].length</code></li>
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<li><code>1 &lt;= n &lt;= 10</code></li>
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<li><code>mat[i][j]</code> and <code>target[i][j]</code> are either <code>0</code> or <code>1</code>.</li>
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</ul>
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## Solutions
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<!-- tabs:start -->
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### **Python3**
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```python
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class Solution:
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def findRotation(self, mat: List[List[int]], target: List[List[int]]) -> bool:
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def rotate(matrix):
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n = len(matrix)
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for i in range(n // 2):
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for j in range(i, n - 1 - i):
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t = matrix[i][j]
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matrix[i][j] = matrix[n - j - 1][i]
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matrix[n - j - 1][i] = matrix[n - i - 1][n - j - 1]
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matrix[n - i - 1][n - j - 1] = matrix[j][n - i - 1]
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matrix[j][n - i - 1] = t
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for _ in range(4):
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if mat == target:
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return True
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rotate(mat)
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return False
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```
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### **Java**
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```java
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class Solution {
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public boolean findRotation(int[][] mat, int[][] target) {
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int times = 4;
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while (times-- > 0) {
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if (equals(mat, target)) {
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return true;
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}
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rotate(mat);
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}
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return false;
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}
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private void rotate(int[][] matrix) {
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int n = matrix.length;
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for (int i = 0; i < n / 2; ++i) {
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for (int j = i; j < n - 1 - i; ++j) {
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int t = matrix[i][j];
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matrix[i][j] = matrix[n - j - 1][i];
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matrix[n - j - 1][i] = matrix[n - i - 1][n - j - 1];
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matrix[n - i - 1][n - j - 1] = matrix[j][n - i - 1];
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matrix[j][n - i - 1] = t;
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}
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}
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}
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private boolean equals(int[][] nums1, int[][] nums2) {
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int n = nums1.length;
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for (int i = 0; i < n; ++i) {
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for (int j = 0; j < n; ++j) {
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if (nums1[i][j] != nums2[i][j]) {
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return false;
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}
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}
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}
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return true;
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}
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}
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```
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### **...**
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```
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```
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<!-- tabs:end -->

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