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 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.
Example:
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4) Output: 7 -> 0 -> 8 Explanation: 342 + 465 = 807.
Solution
Following is my solution, as long as you can deal with the carry well, you can solve the problem easily.
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ class Solution { public ListNode addTwoNumbers(ListNode l1, ListNode l2) { ListNode p1 = l1, p2 = l2; int carry = 0; ListNode result = null; ListNode p = result; for(; p1 != null || p2 != null; ) { int val1 = (p1 != null) ? p1.val : 0; int val2 = (p2 != null) ? p2.val : 0; int sum = val1 + val2 + carry; carry = sum / 10; if(result == null) { result = new ListNode(sum % 10); p = result; } else { p.next = new ListNode(sum % 10); p = p.next; } p1 = p1 != null ? p1.next : p1; p2 = p2 != null ? p2.next : p2; } if(carry != 0) { p.next = new ListNode(carry); } return result; } }
Solution using a dummy head
As you can see in my solution, the leading node of the linked list is a special case and requires an extra case. To make it easier, a dummy head can be used as in the solution provided by Leetcode.
public ListNode addTwoNumbers(ListNode l1, ListNode l2) { ListNode dummyHead = new ListNode(0); ListNode p = l1, q = l2, curr = dummyHead; int carry = 0; while (p != null || q != null) { int x = (p != null) ? p.val : 0; int y = (q != null) ? q.val : 0; int sum = carry + x + y; carry = 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; }