# 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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• Peter

``` public static LinkedList addNumbers(LinkedList l1, LinkedList l2){ LinkedList result = new LinkedList(); int value=0; int carry=0; Node iterator1 = l1.getHead(); Node iterator2 = l2.getHead();```

``` while(iterator1!=null && iterator2 !=null){ value=iterator1.getValue()+iterator2.getValue(); //checking if it is the last node of the list // and if the sum needs another node. if(iterator1.getPointer()==null && value+carry>9){ result.insertLast((value+carry)%10); result.insertLast(1); } else{ result.insertLast((value+carry)%10); if(value+carry>9) carry=1; else carry=0; } iterator1=iterator1.getPointer(); iterator2=iterator2.getPointer(); } return result; } ```

• Peter

int value=0;
int carry=0;
Node iterator1 = l1.getHead();
Node iterator2 = l2.getHead();

while(iterator1!=null && iterator2 !=null){
value=iterator1.getValue()+iterator2.getValue();
//checking if it is the last node of the list
// and if the sum needs another node.
if(iterator1.getPointer()==null && value+carry>9){
result.insertLast((value+carry)%10);
result.insertLast(1);
}
else{
result.insertLast((value+carry)%10);
if(value+carry>9)
carry=1;
else
carry=0;
}
iterator1=iterator1.getPointer();
iterator2=iterator2.getPointer();
}
return result;
}

• Arrow

This might not work in the case when you add 1 and 0. Probably do if (p1.val+p2.val )>9.

• Peeyush Chandel

``` public ListNode addTwoNumbers(ListNode l1, ListNode l2) { int carry = 0; int num = 0; ListNode r = null; ListNode head = null; while (l1 != null || l2 != null) { if (l1 != null && l2 != null) { num = l1.val + l2.val; l1 = l1.next; l2 = l2.next; } else if (l1 != null) { num = l1.val; l1 = l1.next; } else if (l2 != null) { num = l2.val; l2 = l2.next; } if ((num + carry) >= 10) { num = (num + carry) - 10; carry = 1; } else { num = num + carry; carry = 0; } if (r == null) { r = new ListNode(num); head = r; } else { ListNode l = new ListNode(num); r.next = l; r = r.next; }```

``` ```

``` } if (carry != 0) { ListNode l = new ListNode(carry); r.next = l; } return head; } ```

• 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?