Add Two Numbers

To solve this problem, I first visualize the linked lists as stacks of digits in reverse order, similar to how we add numbers manually. I would start by initializing a dummy head node to act as a placeholder for the result list, which keeps my current pointer clean. I also initialize a variable called 'carry' to zero. Next, I iterate through both lists using a while loop that continues as long as either list has nodes or there is a non-zero carry value. Inside the loop, I extract the values from the current nodes of both lists, defaulting to zero if a list is exhausted. I then calculate the total sum by adding these two values along with the current carry. The digit to store in the new node is the sum modulo ten, and the new carry becomes the integer division of the sum by ten. After creating the new node and linking it to my result list, I advance the pointers for both input lists. Once the loop finishes, I return the next node of my dummy head, effectively skipping the placeholder. For example, adding 342 and 465 represented as [2,4,3] and [5,6,4] results in [7,0,8]. This approach ensures we handle different list lengths and final carries correctly in a single pass.