Add Two Numbers Leetcode Solution | C++, Java, Python | Easy Approach

Minimum Cost to Merge Stones
Share:

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.

Add Two Numbers Solutions

Time: O(n)
Space: O(1)

C++

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

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

    return dummy.next;
  }
};

Java

class Solution {
  public int[] twoSum(int[] nums, int target) {
    Map<Integer, Integer> numToIndex = new HashMap<>();

    for (int i = 0; i < nums.length; ++i) {
      if (numToIndex.containsKey(target - nums[i]))
        return new int[] {numToIndex.get(target - nums[i]), i};
      numToIndex.put(nums[i], i);
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;
  }
}

Python

class Solution:
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

Watch Tutorial

Checkout more Solutions here

Leave a Comment

Your email address will not be published. Required fields are marked *

x