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;
}