2.-add-two-numbers.md

2. Add Two Numbers

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 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.

JavaScript

Result

Runtime: 116 ms, faster than 85.61% of JavaScript online submissions for Add Two Numbers.

Memory Usage: 47.2 MB, less than 71.47% of JavaScript online submissions for Add Two Numbers.

/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */

/**
 * @param {ListNode} l1
 * @param {ListNode} l2
 * @return {ListNode}
 */

const addTwoNumbers = (l1, l2) => {
    let current = new ListNode(0);

    const returnList = current;

    let sum, carry = 0;

    while (l1 || l2 || carry > 0) {
        sum = 0;

        if (l1 !== null) {
            sum += l1.val;
            l1 = l1.next;
        }

        if (l2 !== null) {
            sum += l2.val;
            l2 = l2.next;
        }

        sum += carry;

        current.next = new ListNode(sum % 10);
        current = current.next;

        carry = sum > 9 ? 1 : 0;
    }
    return returnList.next;
};

PHP

Result

Runtime: 23 ms, faster than 86.94% of PHP online submissions for Add Two Numbers.

Memory Usage: 19.3 MB, less than 69.59% of PHP online submissions for Add Two Numbers.

class Solution
{
    /**
     * @param ListNode $l1
     * @param ListNode $l2
     * @return ListNode
     */
    function addTwoNumbers(ListNode $l1, ListNode $l2): ?ListNode
    {
        $plus = bcadd($this->getNumber($l1), $this->getNumber($l2));
        $len = strlen($plus);
        $node = null;
        for ($i = 0; $i < $len; $i++) {
            $node = new ListNode($plus[$i], $node);
        }
        return $node;
    }

    function getNumber(ListNode $node): string
    {
        $res = '';
        while ($node->next) {
            $res .= $node->val;
            $node = $node->next;
        }
        return strrev($res . $node->val);
    }
}

Python

Result

Runtime: 69 ms, faster than 94.10% of Python3 online submissions for Add Two Numbers.

Memory Usage: 13.8 MB, less than 86.42% of Python3 online submissions for Add Two Numbers.

class Solution:
    def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]:
        carry = 0
        root = n = ListNode(0)
        while l1 or l2 or carry:
            v1 = v2 = 0
            if l1:
                v1 = l1.val
                l1 = l1.next
            if l2:
                v2 = l2.val
                l2 = l2.next
            carry, val = divmod(v1 + v2 + carry, 10)
            n.next = ListNode(val)
            n = n.next
        return root.next