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[LeetCode 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

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.

Thoughts

  • A dummy head helps with the single linked list problem
  • We can loop two lists until they are all null, note their lengths may be different
  • Carry value is used in the current value calculation
  • Note there may be ending extra carry

Code

Typescript

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

function addTwoNumbers(l1: ListNode | null, l2: ListNode | null): ListNode | null {
    let dummyHead = new ListNode(0);
    let cur = dummyHead;
    let l1Val, l2Val, tempVal, curVal, carry = 0;
    
    while(l1 !== null || l2 !== null) {    
        // handle null node value access
        l1Val = l1 === null ? 0 : l1.val;
        l2Val = l2 === null ? 0 : l2.val;
        // calculate value with the carry
        tempVal = l1Val + l2Val + carry;
        carry = tempVal >= 10 ? 1 : 0;
        curVal = carry ? tempVal - 10 : tempVal
        cur.next = new ListNode(curVal);
        cur = cur.next;
        
        // go to next node if possible
        l1 = l1 === null ? l1 : l1.next;
        l2 = l2 === null ? l2 : l2.next;
    }

    // don't forget the ending carry
    if(carry) {
        cur.next = new ListNode(1)
    }
    
    return dummyHead.next;
};

Reference