feat: add python and java solutions to lcci problem

添加《程序员面试金典》题解：面试题 02.02. 返回倒数第 k 个节点

Author: yanglbme

lcci/面试题 02.02. 返回倒数第 k 个节点/README.md

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+# [面试题 02.02. 返回倒数第 k 个节点](https://leetcode-cn.com/problems/kth-node-from-end-of-list-lcci/)
+
+## 题目描述
+实现一种算法，找出单向链表中倒数第 k 个节点。返回该节点的值。
+
+**注意：** 本题相对原题稍作改动
+
+**示例：**
+
+```
+输入： 1->2->3->4->5 和 k = 2
+输出： 4
+```
+
+**说明：**
+
+- 给定的 k 保证是有效的。
+
+## 解法
+### Python3
+```python
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.next = None
+
+class Solution:
+    def kthToLast(self, head: ListNode, k: int) -> int:
+        p = q = head
+        for _ in range(k):
+            q = q.next
+        while q:
+            q = q.next
+            p = p.next
+        return p.val
+```
+
+### Java
+```java
+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ *     int val;
+ *     ListNode next;
+ *     ListNode(int x) { val = x; }
+ * }
+ */
+class Solution {
+    public int kthToLast(ListNode head, int k) {
+        ListNode p = head, q = head;
+        while (k-- > 0) {
+            q = q.next;
+        }
+        while (q != null) {
+            q = q.next;
+            p = p.next;
+        }
+        return p.val;
+    }
+}
+```
+
+### ...
+```
+
+```

lcci/面试题 02.02. 返回倒数第 k 个节点/Solution.java

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

lcci/面试题 02.02. 返回倒数第 k 个节点/Solution.py

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+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.next = None
+
+class Solution:
+    def kthToLast(self, head: ListNode, k: int) -> int:
+        p = q = head
+        for _ in range(k):
+            q = q.next
+        while q:
+            q = q.next
+            p = p.next
+        return p.val