feat: azl397985856#445 Add python3 solution (azl397985856#270)

* Update 445.add-two-numbers-ii.md

add python3 solution

* Update 445.add-two-numbers-ii.md

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

Author: jojoxiaojing

problems/445.add-two-numbers-ii.md

@@ -37,7 +37,7 @@ Output: 7 -> 8 -> 0 -> 7
 
 ## 代码
 
-- 语言支持：JS，C++
+- 语言支持：JS，C++, Python3
 
 JavaScript Code:
 
@@ -210,3 +210,50 @@ private:
     }
 };
 ```
+
+Python Code:
+
+```python
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.next = None
+
+class Solution:
+    def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
+        def listToStack(l: ListNode) -> list:
+            stack, c = [], l
+            while c:
+                stack.append(c.val)
+                c = c.next
+            return stack
+            
+        # transfer l1 and l2 into stacks
+        stack1, stack2 = listToStack(l1), listToStack(l2)
+            
+        # add stack1 and stack2
+        diff = abs(len(stack1) - len(stack2))
+        stack1 = ([0]*diff + stack1 if len(stack1) < len(stack2) else stack1)
+        stack2 = ([0]*diff + stack2 if len(stack2) < len(stack1) else stack2)
+        stack3 = [x + y for x, y in zip(stack1, stack2)]
+        
+        # calculate carry for each item in stack3 and add one to the item before it
+        carry = 0
+        for i, val in enumerate(stack3[::-1]):
+            index = len(stack3) - i - 1
+            carry, stack3[index] = divmod(val + carry, 10)
+            if carry and index == 0: 
+                stack3 = [1] + stack3
+            elif carry:
+                stack3[index - 1] += 1
+        
+        # transfer stack3 to a linkedList
+        result = ListNode(0)
+        c = result
+        for item in stack3:
+            c.next = ListNode(item)
+            c = c.next
+        
+        return result.next
+```