feat: add python and java solution to lcof problem

添加《剑指 Offer》题解：面试题22. 链表中倒数第k个节点

Author: yanglbme

lcof/README.md

@@ -38,7 +38,11 @@
 │   ├── README.md
 │   ├── Solution.java
 │   └── Solution.py
-└── 面试题11. 旋转数组的最小数字
+├── 面试题11. 旋转数组的最小数字
+│   ├── README.md
+│   ├── Solution.java
+│   └── Solution.py
+└── 面试题22. 链表中倒数第k个节点
     ├── README.md
     ├── Solution.java
     └── Solution.py

lcof/面试题22. 链表中倒数第k个节点/README.md

@@ -0,0 +1,68 @@
+# [面试题22. 链表中倒数第k个节点](https://leetcode-cn.com/problems/lian-biao-zhong-dao-shu-di-kge-jie-dian-lcof/)
+
+## 题目描述
+输入一个链表，输出该链表中倒数第 k 个节点。为了符合大多数人的习惯，本题从 1 开始计数，即链表的尾节点是倒数第 1 个节点。例如，一个链表有 6 个节点，从头节点开始，它们的值依次是 1、2、3、4、5、6。这个链表的倒数第 3 个节点是值为 4 的节点。
+
+
+**示例：**
+
+给定一个链表: 1->2->3->4->5, 和 k = 2.
+
+返回链表 4->5.
+
+## 解法
+### Python3
+```python
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.next = None
+
+class Solution:
+    def getKthFromEnd(self, head: ListNode, k: int) -> ListNode:
+        if not (head or head.next):
+            return head
+        
+        p = q = head
+        for _ in range(k):
+            p = p.next
+        while p:
+            p = p.next
+            q = q.next
+        return q
+
+```
+
+### Java
+```java
+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ *     int val;
+ *     ListNode next;
+ *     ListNode(int x) { val = x; }
+ * }
+ */
+class Solution {
+    public ListNode getKthFromEnd(ListNode head, int k) {
+        if (head == null || head.next == null) {
+            return head;
+        }
+        ListNode p = head, q = head;
+        while (k-- > 0) {
+            p = p.next;
+        }
+        while (p != null) {
+            p = p.next;
+            q = q.next;
+        }
+        return q;
+    }
+}
+```
+
+### ...
+```
+
+```

lcof/面试题22. 链表中倒数第k个节点/Solution.java

@@ -0,0 +1,24 @@
+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ *     int val;
+ *     ListNode next;
+ *     ListNode(int x) { val = x; }
+ * }
+ */
+class Solution {
+    public ListNode getKthFromEnd(ListNode head, int k) {
+        if (head == null || head.next == null) {
+            return head;
+        }
+        ListNode p = head, q = head;
+        while (k-- > 0) {
+            p = p.next;
+        }
+        while (p != null) {
+            p = p.next;
+            q = q.next;
+        }
+        return q;
+    }
+}

lcof/面试题22. 链表中倒数第k个节点/Solution.py

@@ -0,0 +1,18 @@
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.next = None
+
+class Solution:
+    def getKthFromEnd(self, head: ListNode, k: int) -> ListNode:
+        if not (head or head.next):
+            return head
+        
+        p = q = head
+        for _ in range(k):
+            p = p.next
+        while p:
+            p = p.next
+            q = q.next
+        return q