The “Add Two Numbers” leetcode problem is a classic algorithmic challenge frequently encountered in coding interviews. The problem statement goes like this: Given two non-empty linked lists representing two non-negative integers, where each node in the linked list contains a single digit and the digits are stored in reverse order, add the two numbers and return the sum as a linked list. It’s assumed that the two numbers do not contain any leading zeros, except for the number 0 itself.
l1 = [2,4,3]
l2 = [5,6,4]
[7,0,8]
Explanation: 342 + 465 = 807.
To solve this problem, we need to simulate the process of adding two numbers. Since the digits are stored in reverse order, we can traverse the linked lists simultaneously, adding corresponding digits along with any carry from the previous step. The result is a new linked list representing the sum of the two numbers.
Here’s the C# implementation of the solution:
public class AddTwoNumbersSolution
{
public ListNode AddTwoNumbers(ListNode l1, ListNode l2)
{
ListNode result = new ListNode();
ListNode current = result;
int carry = 0;
while (l1 != null || l2 != null || carry != 0)
{
int sum = (l1?.val ?? 0) + (l2?.val ?? 0) + carry;
carry = sum / 10;
current.next = new ListNode(sum % 10);
current = current.next;
l1 = l1?.next;
l2 = l2?.next;
}
return result.next;
}
}
result
linked list to store the sum.current
pointer to keep track of the current node in the result linked list.l1
and l2
) simultaneously, calculating the sum of corresponding digits along with any carry from the previous step.The “Add Two Numbers” problem is an excellent exercise in linked list manipulation and simulating mathematical operations. The solution provides an efficient way to add two numbers represented as linked lists, and the approach can be generalized to similar problems involving linked lists.
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