2.md

2. Add Two Numbers

Description

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

Thinking Process

1. The question is straightforward, the trick is how to create and store the new result
2. Create a virtual head that will not be returned (only as a pointer to the listnode to be returned)
3. Traverse through the two list, create new ListNode only when there is need to do so (l1 != null or l2 != null or carry == 1)
4. Return the next node of the virtual head (where the result really starts)

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
   	int carry = 0;
    ListNode head = new ListNode(0);
    ListNode headCopy = head;

    while(l1 != null || l2 != null || carry == 1){
        int num1 = (l1 == null) ? 0 : l1.val;
        int num2 = (l2 == null) ? 0 : l2.val;
        int sum = num1 + num2 + carry;
        
        if(sum < 10){
            ListNode current = new ListNode(sum);
            headCopy.next = current;
            headCopy = current;
            carry = 0;
        }else{
            ListNode current = new ListNode(sum - 10);
            headCopy.next = current;
            headCopy = current;
            carry = 1;
        }
        
        l1 = (l1 == null) ? null : l1.next;
        l2 = (l2 == null) ? null : l2.next;
    }
    return head.next;
}

Complexity

1. Two linkedlists need to be traversed exactly once, the time complexity is hence O(n)
2. Space complexity is O(n) for storing the result