LeetCode #2: Add Two Numbers (C/C++).

medium

source: https://leetcode.com/problems/add-two-numbers
C/C++ Solution to LeetCode problem 2. Add Two Numbers.

Problem

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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.

Examples

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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

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• 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.

Solution

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The solution is simple

• Two pointers moving at the same time.
• Create a node with value 0 and add the value of the previous two pointers (if they exist).
• As long as any of the pointers have a next node or the sum of the current nodes is > 9, create a new node with the carrying value of the current sum.

Solution:

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        ListNode* result = new ListNode(0);
        ListNode* a = l1;
        ListNode* b = l2;
        ListNode* r = result;

        int carry = 0;
        while (a || b) {
          if (a) {
            r->val += a->val;
            a = a->next;
          }
          if (b) {
            r->val += b->val;
            b = b->next;
          }

          carry = r->val / 10;
          r->val = r->val - (carry * 10);
          if (carry > 0 || a || b) {
            r->next = new ListNode(carry);
            r = r->next;
          }
        }

        return result;
    }
};