I’ve been working through some coding challenges and stumbled upon a pretty interesting problem that has me scratching my head a bit. It involves linked lists representing large numbers, but there’s a twist: the digits are stored in reverse order. So, instead of the typical notation we’re used to, the least significant digit is at the front.
Here’s the scenario: Imagine you’ve got two linked lists, and each one represents a really big number. For instance, if one list is \(2 \rightarrow 4 \rightarrow 3\), that translates to the number 342. The second list could be something like \(5 \rightarrow 6 \rightarrow 4\), which represents 465. The tricky part is that your task is to create a function that adds these two numbers together and outputs a new linked list that also has the result in reverse order.
For example, if we add \(342 + 465\), we expect to get \(807\) in reverse, which would be represented as \(7 \rightarrow 0 \rightarrow 8\) in the new linked list.
Now, here’s where it gets a bit tricky: we need to account for carry-over during the addition. If the sum of two digits exceeds 9, we need to carry over that extra value to the next node. Like when adding \(9\) and \(5\) — we should write down \(4\) and carry over \(1\) to the next digit.
I’m curious how others might approach this. What’s your strategy for tackling this problem? Would you use a straightforward iteration through both lists, keeping track of carries? Or do you think it would be more efficient to store the digits into arrays first and then process the addition?
I’d love to hear your thoughts or even see some sample code if you’ve tackled something similar! How do you handle the potential edge cases, like if one linked list is longer than the other? Let’s brainstorm this together!
To tackle the problem of adding two linked lists representing large numbers in reverse order, the most straightforward approach is to iterate through both lists simultaneously while keeping track of the carry value when the sum of the digits exceeds 9. This method allows you to process the digits pairwise, adding corresponding nodes from both linked lists. Start by initializing a dummy head for the resultant linked list and a variable to maintain the carry. As you traverse the lists, sum the values from the current nodes along with any carry from the previous addition. The resultant digit can be obtained by taking the modulo of the sum by 10, while the new carry can be determined by integer division of the sum by 10. This ensures that you maintain a correct progression of addition without needing to store intermediate results in arrays.
Regarding edge cases, such as when the lengths of the linked lists differ, the iteration should continue as long as there are still nodes in either list. If one list is exhausted before the other, continue adding the remaining digits of the longer list to the result along with any carry. After processing both lists, if there’s still a carry left, append a new node with the value of the carry. This solution is efficient and leverages the properties of linked lists directly, ensuring that it handles large numbers without additional overhead from array operations. Below is a simple implementation of this strategy in Python:
class ListNode: def __init__(self, value=0, next=None): self.value = value self.next = next def addTwoNumbers(l1, l2): dummy_head = ListNode(0) current = dummy_head carry = 0 while l1 or l2 or carry: val1 = (l1.val if l1 else 0) val2 = (l2.val if l2 else 0) total = val1 + val2 + carry carry = total // 10 current.next = ListNode(total % 10) current = current.next if l1: l1 = l1.next if l2: l2 = l2.next return dummy_head.nextSo, this problem sounds kind of tricky but also fun! I’ve been thinking about how to add these two big numbers represented by linked lists where the digits are all flipped around.
First off, I guess I would probably want to just start by looping through both linked lists, adding the digits together. Like, if I’m at the first node of both lists, I would add the values in those nodes. Then if there’s a carry, I’d need to remember that for the next iteration. It seems straightforward to just keep a variable for the carry — like if I add 9 and 5, I’d add 4 to my result list and keep 1 for next time.
What about keeping track of when one list is longer than the other? Hmm. I think I’d just keep adding until I run out of digits in both lists. If one list still has nodes left while the other is finished, I could just keep adding those leftover nodes to my result, right? And just keep carrying if necessary!
Here’s a quick pseudo-code I thought of, just to lay it out:
function addTwoNumbers(l1, l2) { sumList = new List(); // new LinkedList for the result carry = 0; currentNode1 = l1.head; currentNode2 = l2.head; while (currentNode1 != null || currentNode2 != null || carry > 0) { value1 = (currentNode1 != null) ? currentNode1.value : 0; value2 = (currentNode2 != null) ? currentNode2.value : 0; total = value1 + value2 + carry; carry = Math.floor(total / 10); sumList.addNode(total % 10); if (currentNode1 != null) currentNode1 = currentNode1.next; if (currentNode2 != null) currentNode2 = currentNode2.next; } return sumList; // this should give us the final linked list }This would help me to keep things simple while still handling the carries and the different lengths of lists. I’m not sure if there are more efficient ways, like using arrays, but that feels like it might make it more confusing?
What do you all think? Is my approach on the right track, or is there something I’m missing? Would love to hear your thoughts!