[LeetCode] 2. Add Two Numbers

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.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

Thought process:

Use a dummy node to start making the result list. Solve this problem just like solving an addition on paper:

1. Iterate through both lists and pad the shorter list with 0 so that their lengths are equal.
2. Iterate through both lists again and add numbers.
3. If at the end carry = 1, add it to the end of result list.

Solution:

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/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ public class Solution { public ListNode addTwoNumbers(ListNode l1, ListNode l2) { if (l1 == null) { return l2; } if (l2 == null) { return l1; } ListNode dummy = new ListNode(0); ListNode current = dummy; ListNode node1 = l1; ListNode node2 = l2; int carry = 0; while (node1.next != null || node2.next != null) { if (node1.next == null) { node1.next = new ListNode(0); } else if (node2.next == null) { node2.next = new ListNode(0); } node1 = node1.next; node2 = node2.next; } node1 = l1; node2 = l2; while (node1 != null && node2 != null) { int sum = node1.val + node2.val + carry; int digit = sum % 10; carry = sum / 10; current.next = new ListNode(digit); node1 = node1.next; node2 = node2.next; current = current.next; } if (carry > 0) { current.next = new ListNode(1); } return dummy.next; } }

Time complexity: O(n).