Add Two Numbers LeetCode

Table of Contents

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Add Two Numbers

Solution

				
					class Solution {
 public:
  ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
    ListNode dummy(0);
    ListNode* curr = &dummy;
    int carry = 0;

    while (l1 || l2 || carry) {
      if (l1 != nullptr) {
        carry += l1->val;
        l1 = l1->next;
      }
      if (l2 != nullptr) {
        carry += l2->val;
        l2 = l2->next;
      }
      curr->next = new ListNode(carry % 10);
      carry /= 10;
      curr = curr->next;
    }

    return dummy.next;
  }
};

Explanation

The provided code implements a solution to add two numbers represented as linked lists. It uses a dummy node and performs digit-wise addition while keeping track of the carry value. The resulting sum is stored in a new linked list, which is then returned as the output.

Please note that without the complete code and the definition of ListNode, it is difficult to verify the correctness of the code or provide a more accurate explanation.

				
					class Solution {
  public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
    ListNode dummy = new ListNode(0);
    ListNode curr = dummy;
    int carry = 0;

    while (l1 != null || l2 != null || carry > 0) {
      if (l1 != null) {
        carry += l1.val;
        l1 = l1.next;
      }
      if (l2 != null) {
        carry += l2.val;
        l2 = l2.next;
      }
      curr.next = new ListNode(carry % 10);
      carry /= 10;
      curr = curr.next;
    }

    return dummy.next;
  }
}

Explanation

The given code is designed to add two numbers represented as linked lists. It utilizes a loop to iterate through the linked lists, adding the corresponding nodes together while considering any carry-over digits. The result is stored in a new linked list and returned as the output.

To fix the code, we need to define the ListNode class, which was missing in the provided code snippet. The ListNode class represents a node in the linked list and should contain the necessary properties and methods.

				
					class Solution:
  def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
    dummy = ListNode(0)
    curr = dummy
    carry = 0

    while carry or l1 or l2:
      if l1:
        carry += l1.val
        l1 = l1.next
      if l2:
        carry += l2.val
        l2 = l2.next
      curr.next = ListNode(carry % 10)
      carry //= 10
      curr = curr.next

    return dummy.next

Explanation

The given code is an implementation of a solution for adding two numbers represented as linked lists. It defines a class called Solution with a method addTwoNumbers that takes in two parameters, l1 and l2, both representing linked lists.

The code initializes a dummy node, a curr node that references the dummy node, and a carry variable set to 0. It then enters a loop that continues until there is no carry left and both l1 and l2 have been fully traversed.

Within the loop, the code checks if l1 and l2 exist. If they do, it adds the values of the current nodes to the carry variable and moves to the next nodes. It then creates a new node with the value carry % 10 and links it as the next node of the curr node. The carry is updated by dividing it by 10. Finally, curr moves to the next node.

The method returns the next node of the dummy node, which represents the resulting linked list.