LeetCode in TypeScript
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.
Solution
import { ListNode } from '../../com_github_leetcode/listnode'
/*
* 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 {
const dummyHead: ListNode = new ListNode(0)
let p: ListNode | null = l1
let q: ListNode | null = l2
let curr: ListNode | null = dummyHead
let carry: number = 0
while (p !== null || q !== null) {
const x: number = p !== null ? p.val : 0
const y: number = q !== null ? q.val : 0
const sum: number = carry + x + y
carry = Math.floor(sum / 10)
curr.next = new ListNode(sum % 10)
curr = curr.next
if (p !== null) {
p = p.next
}
if (q !== null) {
q = q.next
}
}
if (carry > 0) {
curr.next = new ListNode(carry)
}
return dummyHead.next
}
export { addTwoNumbers }