feat: azl397985856#23 Add python3 solution (azl397985856#271)

* Update 23.merge-k-sorted-lists.md

Add python3 solution to 23.merge-k-sorted-lists

* Update 23.merge-k-sorted-lists.md

Co-authored-by: lucifer <azl397985856@gmail.com>

Author: jojoxiaojing

problems/23.merge-k-sorted-lists.md

@@ -41,6 +41,10 @@ https://leetcode-cn.com/problems/merge-k-sorted-lists/description/
 
 ## 代码
 
+代码支持 JavaScript， Python3
+
+JavaScript Code：
+
 ```js
 /*
  * @lc app=leetcode id=23 lang=javascript
@@ -49,27 +53,6 @@ https://leetcode-cn.com/problems/merge-k-sorted-lists/description/
  *
  * https://leetcode.com/problems/merge-k-sorted-lists/description/
  *
- * algorithms
- * Hard (33.14%)
- * Total Accepted:    373.7K
- * Total Submissions: 1.1M
- * Testcase Example:  '[[1,4,5],[1,3,4],[2,6]]'
- *
- * Merge k sorted linked lists and return it as one sorted list. Analyze and
- * describe its complexity.
- *
- * Example:
- *
- *
- * Input:
- * [
- * 1->4->5,
- * 1->3->4,
- * 2->6
- * ]
- * Output: 1->1->2->3->4->4->5->6
- *
- *
  */
 function mergeTwoLists(l1, l2) {
   const dummyHead = {};
@@ -129,6 +112,51 @@ var mergeKLists = function(lists) {
 };
 ```
 
+Python3 Code：
+
+``` python
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.next = None
+
+class Solution:
+    def mergeKLists(self, lists: List[ListNode]) -> ListNode:
+        n = len(lists)
+        
+        # basic cases
+        if lenth == 0: return None
+        if lenth == 1: return lists[0]
+        if lenth == 2: return self.mergeTwoLists(lists[0], lists[1])
+        
+        # divide and conqure if not basic cases
+        mid = n // 2
+        return self.mergeTwoLists(self.mergeKLists(lists[:mid]), self.mergeKLists(lists[mid:n]))
+    
+    
+    def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
+        res = ListNode(0)
+        c1, c2, c3 = l1, l2, res
+        while c1 or c2:
+            if c1 and c2:
+                if c1.val < c2.val:
+                    c3.next = ListNode(c1.val)
+                    c1 = c1.next
+                else:
+                    c3.next = ListNode(c2.val)
+                    c2 = c2.next
+                c3 = c3.next
+            elif c1:
+                c3.next = c1
+                break
+            else:
+                c3.next = c2
+                break
+            
+        return res.next
+```
+
 ## 相关题目
 
 -[88.merge-sorted-array](./88.merge-sorted-array.md)