Add Two Numbers

Problem

You are given two non-empty linked lists representing two non-negative integers. The most significant digit comes first 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.

Follow up: What if you cannot modify the input lists? In other words, reversing the lists is not allowed.

Example:

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

Solution

public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        if (l1 == null) return l2;
        if (l2 == null) return l1;

        Stack<Integer> s1 = new Stack<>(), s2 = new Stack<>();
        while (l1 != null) {
            s1.push(l1.val);
            l1 = l1.next;
        }
        while (l2 != null) {
            s2.push(l2.val);
            l2 = l2.next;
        }

        ListNode end = new ListNode(0);
        int sum = 0;
        while (!s1.isEmpty() || !s2.isEmpty()) {
            if (!s1.isEmpty()) sum += s1.pop();
            if (!s2.isEmpty()) sum += s2.pop();
            end.val = sum % 10;
            ListNode head = new ListNode(sum / 10);
            head.next = end;
            end = head;
            sum /= 10;
        }
        return end.val == 0 ? end.next: end;
    }
}

Analysis

We use stacks to get val from linked list in reversed order
Then we set end.val as the sum of two numbers from stacks
Then we create a new head with value sum / 10
Then connect head.next = end and move end to the head
At the end, we need to check end.val == 0 because as long as last sum is not 10 -> 10 / 10 = 1