2. Add Two Numbers

• Difficulty: Medium

• Topics: Linked List, Math

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Problem:

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 contain a single digit. Add the two numbers and return it as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Example:

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

Solutions:

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        ListNode* head = new ListNode(0);
        ListNode* cur = head;
        int carry = 0;

        while (l1 != NULL || l2 != NULL) {
            int val1 = 0;
            int val2 = 0;
            if (l1) {
                val1 = l1->val;
                l1 = l1->next;
            } 
            if (l2) {
                val2 = l2->val;
                l2 = l2->next;
            }

            int sum = val1 + val2 + carry;
            carry = sum/10;
            cur->next = new ListNode(sum%10);
            cur = cur->next;
        }

        if (carry == 1) {
            cur->next = new ListNode(1);
        }

        return head->next;
    }
};

Another More Solution

This solution breaks the while loop earlier to avoid unnecessary list traversal.

class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        ListNode* head = new ListNode(0);
        ListNode* cur = head;
        int carry = 0;

        while (carry == 1 || (l1 != NULL && l2 != NULL)) {
            int val1 = 0;
            int val2 = 0;
            if (l1) {
                val1 = l1->val;
                l1 = l1->next;
            } 
            if (l2) {
                val2 = l2->val;
                l2 = l2->next;
            }

            int sum = val1 + val2 + carry;
            carry = sum/10;
            cur->next = new ListNode(sum%10);
            cur = cur->next;
        }

        if (l1 != NULL) {
            cur->next = l1;
        } else {
            cur->next = l2;
        }

        return head->next;
    }
};