Leetcode 2. Add Two Numbers

You are given two linked lists representing two non-negative numbers. 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.

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

Analysis:

• Initialize current node to dummy head of the returning list.
• Initialize carry to $0$.
• Initialize $p$ and $q$ to head of $l1$ and $l2$ respectively.
• Loop through lists $l1$ and $l2$ until you reach both ends.
  • Set $x$ to node $p$‘s value. If $p$ has reached the end of $l1$, set to $0$.
  • Set $y$ to node $q$‘s value. If $q$ has reached the end of $l2$, set to $0$.
  • Set $sum=x+y+carry$.
  • Update $carry=sum/10$.
  • Create a new node with the digit value of $(summod10)$ and set it to current node’s next, then advance current node to next.
  • Advance both $p$ and $q$.
• Check if $carry=1$, if so append a new node with digit $1$ to the returning list.
• Return dummy head’s next node.

Time complexity: O(max(m, n))

Space complexity: O(max(m, n))

Code is below:

/**
 * 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) {
        // O(n) O(1)
        
        if (l1 == null || l2 == null) {
            return null;
        }
        
        ListNode dummy = new ListNode(0);
        ListNode cur = dummy;
        
        int sum = 0;
        int carry = 0;
        
        while (l1 != null || l2 != null) {
            int num1 = l1 == null ? 0 : l1.val;
            int num2 = l2 == null ? 0 : l2.val;
            
            sum = num1 + num2 + carry;
            
            // update the next node
            cur.next = new ListNode(sum % 10);
            
            cur = cur.next;
            carry = sum / 10;
            
            l1 = l1 == null ? null : l1.next;
            l2 = l2 == null ? null : l2.next;
        }
        
        // Check the last corner case
        if (carry != 0) {
            cur.next = new ListNode(carry);
        }
        
        return dummy.next;
    }
}