---
# System prepended metadata

title: Add Two Numbers (not finish)
tags: [data_structure_python]

---

---
tags: data_structure_python
---

# Add Two Numbers (not finish)

You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

**Example:**

```
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.
```

# Solution 


### Solution 1: Iterative 1 (clean the 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:        
        # Time complexity: O(max(len(l1), len(l2))).
        # Space complexity: O(max(len(l1), len(l2))).
        
        p1, p2 = l1, l2
        node = ListNode(None)
        pNode = node
        carry = 0
        
        while p1 != None and p2 != None:
            x = p1.val + p2.val + carry

            if x > 9:
                carry = x // 10
                x = x % 10
            else:
                carry = 0
                
            pNode.next = ListNode(x)
            pNode = pNode.next
            
            p1 = p1.next
            p2 = p2.next
        
        while p1 != None: # len(p1) > len(p2).
            x = p1.val + carry
            
            if x > 9:
                carry = x // 10
                x = x % 10
            else:
                carry = 0
                
            pNode.next = ListNode(x)
            pNode = pNode.next
            p1 = p1.next
            
        while p2 != None: # len(p2) > len(p1).
            x = p2.val + carry
            
            if x > 9:
                carry = x // 10
                x = x % 10
            else:
                carry = 0
                
            pNode.next = ListNode(x)
            pNode = pNode.next
            p2 = p2.next
        
        if carry != 0:
            pNode.next = ListNode(carry)
        
        return node.next
            
```
### Solution 2: Iterative 2 (clean the code)

### Solution 3: Recursive