2. Add Two Numbers LeetCode Solution

In this guide we will provide 2. Add Two Numbers LeetCode Solution with best time and space complexity. The solution to Add Two Numbers problem is provided in various programming languages like C++, Java and python. This will be helpful for you if you are preparing for placements, hackathon, interviews or practice purposes. The solutions provided here are very easy to follow and with detailed explanations.

Table of Contents

2. Add Two Numbers LeetCode Solution image

Problem Statement of Add Two Numbers

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.

Complexity Analysis

  • Time Complexity: O(n)
  • Space Complexity: O(1)

2. Add Two Numbers LeetCode Solution in C++

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

    while (l1 != nullptr || l2 != nullptr || carry > 0) {
      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;
  }
};
/* code provided by PROGIEZ */

2. Add Two Numbers LeetCode Solution in Java

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;
  }
}
// code provided by PROGIEZ

2. Add Two Numbers LeetCode Solution in Python

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
#code by PROGIEZ

Additional Resources

See also  18. 4Sum LeetCode Solution

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