feat: add python and java solution to lcof problem

添加《剑指 Offer》题解：面试题18. 删除链表的节点

Author: yanglbme

lcof/README.md

@@ -46,7 +46,11 @@
 │   ├── README.md
 │   ├── Solution.java
 │   └── Solution.py
-└── 面试题24. 反转链表
+├── 面试题24. 反转链表
+│   ├── README.md
+│   ├── Solution.java
+│   └── Solution.py
+└── 面试题25. 合并两个排序的链表
     ├── README.md
     ├── Solution.java
     └── Solution.py

lcof/面试题18. 删除链表的节点/README.md

@@ -0,0 +1,88 @@
+# [面试题18. 删除链表的节点](https://leetcode-cn.com/problems/shan-chu-lian-biao-de-jie-dian-lcof/)
+
+## 题目描述
+给定单向链表的头指针和一个要删除的节点的值，定义一个函数删除该节点。
+
+返回删除后的链表的头节点。
+
+注意：此题对比原题有改动
+
+**示例 1:**
+
+```
+输入: head = [4,5,1,9], val = 5
+输出: [4,1,9]
+解释: 给定你链表中值为 5 的第二个节点，那么在调用了你的函数之后，该链表应变为 4 -> 1 -> 9.
+```
+
+**示例 2:**
+
+```
+输入: head = [4,5,1,9], val = 1
+输出: [4,5,9]
+解释: 给定你链表中值为 1 的第三个节点，那么在调用了你的函数之后，该链表应变为 4 -> 5 -> 9.
+```
+
+**说明：**
+
+- 题目保证链表中节点的值互不相同
+- 若使用 C 或 C++ 语言，你不需要 `free` 或 `delete` 被删除的节点
+
+## 解法
+### Python3
+```python
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.next = None
+
+class Solution:
+    def deleteNode(self, head: ListNode, val: int) -> ListNode:
+        pre = ListNode(0)
+        pre.next = head
+        dummy = pre
+        p = head
+        while p:
+            if p.val == val:
+                pre.next = p.next
+                break
+            else:
+                pre, p = p, p.next
+        return dummy.next
+
+```
+
+### Java
+```java
+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ *     int val;
+ *     ListNode next;
+ *     ListNode(int x) { val = x; }
+ * }
+ */
+class Solution {
+    public ListNode deleteNode(ListNode head, int val) {
+        ListNode pre = new ListNode(0);
+        pre.next = head;
+        ListNode dummy = pre, p = head;
+        while (p != null) {
+            if (p.val == val) {
+                pre.next = p.next;
+                break;
+            } else {
+                pre = p;
+                p = p.next;
+            }
+        }
+        return dummy.next;
+    }
+}
+```
+
+### ...
+```
+
+```

lcof/面试题18. 删除链表的节点/Solution.java

@@ -0,0 +1,21 @@
+/**
+ * Definition for singly-linked list. public class ListNode { int val; ListNode
+ * next; ListNode(int x) { val = x; } }
+ */
+class Solution {
+    public ListNode deleteNode(ListNode head, int val) {
+        ListNode pre = new ListNode(0);
+        pre.next = head;
+        ListNode dummy = pre, p = head;
+        while (p != null) {
+            if (p.val == val) {
+                pre.next = p.next;
+                break;
+            } else {
+                pre = p;
+                p = p.next;
+            }
+        }
+        return dummy.next;
+    }
+}

lcof/面试题18. 删除链表的节点/Solution.py

@@ -0,0 +1,19 @@
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.next = None
+
+class Solution:
+    def deleteNode(self, head: ListNode, val: int) -> ListNode:
+        pre = ListNode(0)
+        pre.next = head
+        dummy = pre
+        p = head
+        while p:
+            if p.val == val:
+                pre.next = p.next
+                break
+            else:
+                pre, p = p, p.next
+        return dummy.next