Leetcode – Add Two Numbers (Java)

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

Java Solution

public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
       int carry =0;
 
        ListNode newHead = new ListNode(0);
        ListNode p1 = l1, p2 = l2, p3=newHead;
 
        while(p1 != null || p2 != null){
            if(p1 != null){
                carry += p1.val;
                p1 = p1.next;
            }
 
            if(p2 != null){
                carry += p2.val;
                p2 = p2.next;
            }
 
            p3.next = new ListNode(carry%10);
            p3 = p3.next;
            carry /= 10;
        }
 
        if(carry==1) 
            p3.next=new ListNode(1);
 
        return newHead.next;
    }
}

What if the digits are stored in regular order instead of reversed order?

Answer: We can simple reverse the list, calculate the result, and reverse the result.

Category >> Algorithms  
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  • Udaydeep Thota

    You could do it, but its not a good practice to navigate the list with the head pointers.

  • Udaydeep Thota

    Works for all the cases, you are just changing the order of numbers and adding it in the same way. Eg: 490 +829 = 1319. Using this method gives 094+928 = 9131 which is correct (Reverse of 9131 is 1319)

  • Milan

    A slightly simpler version,

    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
    int reminder = 0, sum = 0, nextDigit = 0;
    ListNode dummyNode = new ListNode(0);
    ListNode T = dummyNode;
    while (l1 !=null && l2 != null){
    sum = l1.val+l2.val+reminder;
    nextDigit = sum > 9 ? sum-10 : sum;
    reminder = sum > 9 ? 1 : 0;
    T.next = new ListNode(nextDigit);
    l1=l1.next;
    l2=l2.next;
    T=T.next;
    }
    while (l1 !=null){
    sum = l1.val+reminder;
    nextDigit = sum > 9 ? sum-10 : sum;
    reminder = sum > 9 ? 1 : 0;
    l1=l1.next;
    T.next = new ListNode(nextDigit);
    T=T.next;
    }
    while (l2 !=null){
    sum = l2.val+reminder;
    nextDigit = sum > 9 ? sum-10 : sum;
    reminder = sum > 9 ? 1 : 0;
    l2=l2.next;
    T.next = new ListNode(nextDigit);
    T=T.next;
    }
    if (reminder != 0 )
    T.next = new ListNode(reminder);
    return dummyNode.next;
    }

  • Rahul

    Thanks for the code.. Logically its correct. The only concern is about adding a 4 digit number to a 3 digit number in regular order

  • Jayce Kim

    I am also curious about that. Did you figure it out?

  • random

    Bunch of looser have you guys tested it once ?
    it wasted my 50 min of time.

  • Zach

    Doesn’t work if first two numbers added together are >= 10 does it?

  • Wonderer

    I wonder, why do you need a new name for l1 and l2 and don’t just use l1 and l2?

  • In this case an additional ListNode is added at the end of resulting sum. The following excerpt solves the case:

    if(carry==1)
    p3.next=new ListNode(1);

  • Mohit

    What if you get 4 digits after adding two 3-digit numbers?