Interview Question Add Two Numbers

Solution by Fatih KABAKCI

Problem #2- Add Two Numbers

Description: 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.

Difficulty: Medium

Solution in Java

/**
 * 
 * @author fkabakci
 * Problem Description: 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.
 * 
 * Examples:
 * 
 * Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
   Output: 7 -> 0 -> 8
   
   Input: (0 -> 1) + (0 -> 1 -> 2)
   Output: 0 -> 2 -> 2
   
   Input: () + (0 -> 1)
   Output: 0 -> 1
   
   Input: (9 -> 9) + (1)
   Output: 0 -> 0 -> 1
   
   Input: (9 -> 9) + (9 -> 9)
   Output: 8 -> 9 -> 1
   
   Solution:
   
   First, annex the carry value too into the sum, which is collection of two digits in the same order.
   Then, update new carry value until end of the node.
 */

class ListNode {
	int val;
	ListNode next;

	public ListNode(int x) {
		val = x;
	}

	public ListNode(int... vals) {
		if (vals.length > 1) {
			ListNode t = this;
			int i;
			for (i = 0; i < vals.length - 1; i++) {
				t.val = vals[i];
				t.next = new ListNode(0);
				t = t.next;
			}
			t.val = vals[i];
		}
	}

	public void print() {
		ListNode t = this;
		while (t != null) {
			System.out.print(t.val + "->");
			t = t.next;
		}
		System.out.println();
	}
}

public class AddTwoNumbers {
	public static ListNode addTwoNumbers(ListNode l1, ListNode l2) {
		ListNode l3 = new ListNode(0);
		ListNode tl1 = l1, tl2 = l2, tl3 = l3;

		int carry = 0;
		while (tl1 != null || tl2 != null) {
			int digit1 = (tl1 != null) ? tl1.val : 0;
			int digit2 = (tl2 != null) ? tl2.val : 0;
			int sum = carry + digit1 + digit2;
			carry = sum / 10;
			tl3.next = new ListNode(sum % 10);
			tl3 = tl3.next;
			if (tl1 != null)
				tl1 = tl1.next;
			if (tl2 != null)
				tl2 = tl2.next;
		}

		if (carry == 1) {
			tl3.next = new ListNode(carry);
		}
		return l3.next;
	}

	public static void main(String[] args) {

		ListNode l1 = new ListNode(2, 4, 3);
		ListNode l2 = new ListNode(5, 6, 4);
		ListNode l3 = addTwoNumbers(l1, l2);
		l3.print();

		ListNode l21 = new ListNode(0, 1);
		ListNode l22 = new ListNode(0, 1, 2);
		ListNode l23 = addTwoNumbers(l21, l22);
		l23.print();

		ListNode l31 = new ListNode();
		ListNode l32 = new ListNode(0, 1);
		ListNode l33 = addTwoNumbers(l31, l32);
		l33.print();

		ListNode l41 = new ListNode(9, 9);
		ListNode l42 = new ListNode(1);
		ListNode l43 = addTwoNumbers(l41, l42);
		l43.print();

		ListNode l51 = new ListNode(9, 9);
		ListNode l52 = new ListNode(9, 9);
		ListNode l53 = addTwoNumbers(l51, l52);
		l53.print();
	}
}

We basically annex the carry value too into the sum, which is collection of two digits in the same order. Then, we update new carry value. Otherwise, we only add sum of two digits into the node.

The algorithm time complexity is O(max(l1, l2)) because whichever is larger than the other one, we travel through the loop as many times as number of elements in larger linked list. Algorithm space complexity is also same as time. We also create nodes as many as number of loop cycle.