Linked list arithmetic

Question

I am practicing for interviews and I tried solving the problem on my own and got a solution. I was wondering how can I improve the performance and memory on this program?

The problem performs addition on linked list nodes. The nodes are 1s->10s->100s->.... The goal of this program is to perform addition on two of these linked lists. The output should be a linked list of the same format.

Example:

  1 -> 4 -> 3 
+ 1 -> 5 -> 9 -> 2 

Linked list output:

2 -> 9 -> 2 -> 3

Here is my code:

public class prob2_5 {
    /*
     * You have two numbers represented by a linked list, where each node contains a single digit.
     * The digits are stored in reverse order, such that the 1's digit is at the head of the list.
     * Write a function that adds the two numbers and returns the sum as a linked list
     * 
     */

    public static LinkedListNode addLists(LinkedListNode l1, LinkedListNode l2) {
        int l1sum = 0;
        int l2sum = 0;
        int multiplier = 1;

        while(l1 != null) {
            l1sum = l1sum + l1.value*multiplier;
            multiplier = multiplier*10;
            l1 = l1.next;
        }
        multiplier = 1;
        while(l2 != null) {
            l2sum = l2sum + l2.value*multiplier;
            multiplier = multiplier*10;
            l2 = l2.next;
        }

        int total = l2sum + l1sum;

        char[] totalnum = (total + "").toCharArray();
        int len = totalnum.length;
        LinkedListNode tail = new LinkedListNode(totalnum[len - 1] - '0');
        LinkedListNode newNode;
        LinkedListNode prevNode = tail;
        System.out.print(totalnum);

        for (int i = len - 2; i >= 0; i--) {            
            newNode = new LinkedListNode(totalnum[i] - '0');
            prevNode.next = newNode;        
            prevNode = newNode; 
        }
        return tail;
    }   
}

Here is my linked list class:

public class LinkedListNode {
    LinkedListNode next;
    int value;
    public LinkedListNode(int value) {
        this.next = null;
        this.value = value;
    }
}

Answer 1

I believe you are missing the point of the exercise - instead of 'decoding' the input numbers, and then 're-encoding' the output number - you are supposed to iteratively get to the answer by going over the nodes of both lists:

public static LinkedListNode addLists(LinkedListNode l1, LinkedListNode l2) {
  return addLists(0, l1, l2);
}

private static LinkedListNode addLists(int carryOver, LinkedListNode l1, LinkedListNode l2) {
  // stop conditions
  if (l1 == null && l2 == null && carryOver == 0) {
    return null;
  }
  if (l1 == null) {
    l1 = new LinkedListNode(0);
  }
  if (l2 == null) {
    l2 = new LinkedListNode(0);
  }

  // iteration
  int addedValue = l1.value + l2.value + carryOver;
  carryOver = 0;

  if (addedValue >= 10) {
    addedValue -= 10;
    carryOver = 1;
  }

  l3 = new LinkedListNode(addedValue);

  // recursion
  l3.next = addLists(carryOver, l1.next, l2.next);

  return l3;
}

Answer 2

Uri has a good answer, this answer is just to add a second spin on the problem....

Recursion is a good tool to have, but it is not necessarily the best tool for all occasions. In this instance, it's a toss-up.... but the iterative solution to this problem is perhaps a simpler thing to read..... and should be considered. Additionally, in many cases an iterative solution will outperform a recursive solution. In this case, again, it is a toss-up because the lists will be so short.... but, if the list was long, the recursive approach will fail with a stack-overflow... the iterative approach will keep trucking though.

public static final LinkedListNode add(LinkedListNode a, LinkedListNode b) {
    // this pointer points to the result, it is not the actual result.
    final LinkedListNode pointer = new LinkedListNode(0);

    int carry = 0;
    LinkedListNode cursor = pointer;
    while (a != null || b != null) {
        int digitsum = carry;
        if (a != null) {
            digitsum += a.value;
            a = a.next;
        }
        if (b != null) {
            digitsum += b.value;
            b = b.next;
        }
        cursor.next = new LinkedListNode(digitsum % 10);
        carry = digitsum / 10;
        cursor = cursor.next;
    }
    if (carry != 0) {
        cursor.next = new LinkedListNode(carry);
    }

    // don't return the dummy pointer, return the actual result
    return pointer.next;
}

Some side-notes:

• the variables l1 and l2 are not great names....
• The LinkedListNode should probably have a final value, and there should be getters for the value and next, and a setter for the next.