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2. Add Two Numbers (Medium)

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.

Companies:
Amazon, Microsoft, Facebook, Bloomberg, Adobe, Apple, Google, Yahoo, Uber, Capital One, JPMorgan, Huawei

Related Topics:
Linked List, Math, Recursion

Similar Questions:

• Multiply Strings (Medium)
• Add Binary (Easy)
• Sum of Two Integers (Medium)
• Add Strings (Easy)
• Add Two Numbers II (Medium)
• Add to Array-Form of Integer (Easy)
• Add Two Polynomials Represented as Linked Lists (Medium)

Solution 1.

// OJ: https://leetcode.com/problems/add-two-numbers/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    ListNode* addTwoNumbers(ListNode* a, ListNode* b) {
        int carry = 0;
        ListNode dummy, *tail = &dummy;
        while (a || b || carry) {
            if (a) {
                carry += a->val;
                a = a->next;
            }
            if (b) {
                carry += b->val;
                b = b->next;
            }
            tail->next = new ListNode(carry % 10);
            tail = tail->next;
            carry /= 10;
        }
        return dummy.next;
    }
};