2 Add Two Numbers - Medium

Problem:

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 contains a single digit. Add the two numbers and return the sum as a linked list.

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

Example 1:

Input: l1 = [2,4,3], l2 = [5,6,4] Output: [7,0,8] Explanation: 342 + 465 = 807.

Example 2:

Input: l1 = [0], l2 = [0] Output: [0]

Example 3:

Input: l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9] Output: [8,9,9,9,0,0,0,1]

Constraints:

• The number of nodes in each linked list is in the range [1, 100].
• 0 <= Node.val <= 9
• It is guaranteed that the list represents a number that does not have leading zeros.

Problem Analysis:

1. Initialize: Create a dummy node and a current pointer (cur) to keep track of the current position in the result linked list.
2. Traverse and Add: Iterate through both linked lists (l1 and l2) and add corresponding node values along with any carry from the previous addition.
3. Update Carry and Value: Calculate the sum (val), update the carry, and create a new node with the result value. Move the current pointer to the next position.
4. Handle Unequal Lengths: Continue the process until you reach the end of both lists. If one list is shorter than the other, use 0 for missing values.
5. Return Result: Return the next node of the dummy node, which contains the result linked list.

Be Careful of Edge Cases

1. Carry at the End: If there is a carry after processing all the nodes, an additional node needs to be added to represent the carry.

2. Different Lengths: The code handles cases where l1 and l2 have different lengths by assigning 0 to the value of the missing node in the shorter list.

Complexity

• Time Complexity: O(max(N, M)), where N and M are the lengths of the input linked lists l1 and l2. The algorithm processes each node once.
• Space Complexity: O(max(N, M)), the space required for the output linked list.

Solutions:

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]:
        dummy = ListNode()
        cur = dummy
        carry = 0

        # edge case where carry exist but no existing node in front eg. 7 + 8
        while l1 or l2 or carry:
            # init empty node for unequal pair
            v1 = l1.val if l1 else 0
            v2 = l2.val if l2 else 0

            val = v1 + v2 + carry

            carry = val // 10
            val = val % 10
            cur.next = ListNode(val)

            # increment
            cur = cur.next
            l1 = l1.next if l1 else None
            l2 = l2.next if l2 else None
        
        return dummy.next
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