feat: add python and java solution to lcof problem

添加《剑指Offer》题解：面试题25. 合并两个排序的链表

Author: yanglbme

lcof/面试题25. 合并两个排序的链表/README.md

@@ -0,0 +1,91 @@
+# [面试题25. 合并两个排序的链表](https://leetcode-cn.com/problems/he-bing-liang-ge-pai-xu-de-lian-biao-lcof/)
+
+## 题目描述
+输入两个递增排序的链表，合并这两个链表并使新链表中的节点仍然是递增排序的。
+
+**示例1：**
+
+```
+输入：1->2->4, 1->3->4
+输出：1->1->2->3->4->4
+```
+
+**限制：**
+
+- `0 <= 链表长度 <= 1000`
+
+## 解法
+### Python3
+```python
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.next = None
+
+class Solution:
+    def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
+        if l1 is None or l2 is None:
+            return l1 or l2
+        head = ListNode(0)
+        p = head
+        while l1 and l2:
+            if l1.val < l2.val:
+                t = l1.next
+                p.next = l1
+                p = l1
+                l1 = t
+            else:
+                t = l2.next
+                p.next = l2
+                p = l2
+                l2 = t
+
+        p.next = l1 or l2
+        return head.next
+```
+
+### Java
+```java
+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ *     int val;
+ *     ListNode next;
+ *     ListNode(int x) { val = x; }
+ * }
+ */
+class Solution {
+    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
+        if (l1 == null) {
+            return l2;
+        }
+        if (l2 == null) {
+            return l1;
+        }
+
+        ListNode head = new ListNode(0);
+        ListNode p = head;
+        while (l1 != null && l2 != null) {
+            if (l1.val < l2.val) {
+                ListNode t = l1.next;
+                p.next = l1;
+                p = l1;
+                l1 = t;
+            } else {
+                ListNode t = l2.next;
+                p.next = l2;
+                p = l2;
+                l2 = t;
+            }
+        }
+        p.next = l1 == null ? l2 : l1;
+        return head.next;
+    }
+}
+```
+
+### ...
+```
+
+```

lcof/面试题25. 合并两个排序的链表/Solution.java

@@ -0,0 +1,36 @@
+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ *     int val;
+ *     ListNode next;
+ *     ListNode(int x) { val = x; }
+ * }
+ */
+class Solution {
+    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
+        if (l1 == null) {
+            return l2;
+        }
+        if (l2 == null) {
+            return l1;
+        }
+
+        ListNode head = new ListNode(0);
+        ListNode p = head;
+        while (l1 != null && l2 != null) {
+            if (l1.val < l2.val) {
+                ListNode t = l1.next;
+                p.next = l1;
+                p = l1;
+                l1 = t;
+            } else {
+                ListNode t = l2.next;
+                p.next = l2;
+                p = l2;
+                l2 = t;
+            }
+        }
+        p.next = l1 == null ? l2 : l1;
+        return head.next;
+    }
+}

lcof/面试题25. 合并两个排序的链表/Solution.py

@@ -0,0 +1,26 @@
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.next = None
+
+class Solution:
+    def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
+        if l1 is None or l2 is None:
+            return l1 or l2
+        head = ListNode(0)
+        p = head
+        while l1 and l2:
+            if l1.val < l2.val:
+                t = l1.next
+                p.next = l1
+                p = l1
+                l1 = t
+            else:
+                t = l2.next
+                p.next = l2
+                p = l2
+                l2 = t
+
+        p.next = l1 or l2
+        return head.next