# 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

```/* 2 -> 4 -> 3 5 -> 6 -> 4 7 0 8 */ public ListNode addTwoNumbers(ListNode l1, ListNode l2) { ListNode fake = new ListNode(0); ListNode p = fake;   ListNode p1 = l1; ListNode p2 = l2;   int carry = 0; while(p1!=null || p2!=null){ int sum = carry; if(p1!=null){ sum += p1.val; p1 = p1.next; }   if(p2!=null){ sum += p2.val; p2 = p2.next; }   if(sum>9){ carry=1; sum = sum-10; }else{ carry = 0; }   ListNode l = new ListNode(sum); p.next = l; p = p.next; }   //don't forget check the carry value at the end if(carry > 0){ ListNode l = new ListNode(carry); p.next = l; } return fake.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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• yvcvenkat 007

` `

``` package com.developerslike; import java.util.LinkedList; public class LinkedListSUM { public static void main(String[] args) { LinkedList linkedList1 = new LinkedList(); LinkedList linkedList2 = new LinkedList(); LinkedList linkedList3 = new LinkedList(); linkedList1.add(2); linkedList1.add(4); linkedList1.add(3); linkedList2.add(5); linkedList2.add(6); linkedList2.add(4); linkedList3 = addTwoLists(linkedList1,linkedList2); linkedList3.forEach(a -> System.out.print(" "+a)); } private static LinkedList addTwoLists(LinkedList l1, LinkedList l2) { int carry =0; LinkedList l3 = new LinkedList(); for(int i=l1.size()-1;i>=0;i--) { int parseInt = Integer.parseInt(l1.get(i).toString()); int parseInt2 = Integer.parseInt(l2.get(i).toString()); int sum =parseInt + parseInt2+carry; int value=sum%10; carry =sum/10; l3.add(value); } return l3; } } ```

• VIVEK SINGH

``` struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2){ int len1=0,len2=0,sum,rem=0; struct ListNode *t1=l1,*t2=l2; while(t1 || t2) { if(t1) { len1++; t1=t1->next; } if(t2) { len2++; t2=t2->next; } } t1=l1,t2=l2; if(len1>len2) { while(t2->next) { t2=t2->next; } while(len1>len2) { struct ListNode* new = (struct ListNode*)malloc(sizeof(struct ListNode)); new->val=0; new->next=NULL; t2->next=new; t2=t2->next; len2++; } } if(len1next) { t1=t1->next; } while(len1val=0; new->next=NULL; t1->next=new; t1=t1->next; len1++; } } t1=l1,t2=l2; while(t1) { sum=t1->val+t2->val+rem; t2->val=sum%10; rem=sum/10; t1=t1->next; t2=t2->next; } if(rem) { t2=l2; while(t2->next) { t2=t2->next; } struct ListNode* new = (struct ListNode*)malloc(sizeof(struct ListNode)); new->val=rem; new->next=NULL; t2->next=new; } return l2; } ```

• Ayush Sharma

What if instead of initializing p and q (as l1 and l2 respectively) we directly use provided l1 and l2 instead? Would it create any difference?

• Bonsai

Right. The first node can be detached, by fake.next = null. But it does not affect the correctness of the code.

``` ListNode result = fake.next; fake.next = null;```

``` ```

• Chaitanya GOPIREDDY

First node in the solution will always be 0.

• Marina

``` package structures;```

``` public class AddTwoNumbers { public LinkedList calculate(LinkedList first, LinkedList second) { int carryOver = 0; LinkedList result = new LinkedList(); Node firstCurrent = first.getHead(); Node secondCurrent = second.getHead(); while (firstCurrent != null && secondCurrent != null) { int res = firstCurrent.getVal() + secondCurrent.getVal() + carryOver; firstCurrent = firstCurrent.getNext(); secondCurrent = secondCurrent.getNext(); result.add(res % 10); carryOver = carry(res); } while (firstCurrent != null) { int res = carryOver + firstCurrent.getVal(); result.add(res% 10); carryOver = carry(res); } while (secondCurrent != null) { int res = carryOver + secondCurrent.getVal(); result.add(res% 10); carryOver = carry(res); } return result; } private int carry(int res) { int carryOver; if (res > 9) { carryOver = 1; } else { carryOver = 0; } return carryOver; } } ```

• Работа в интернете

• Kanak Lata

int carry=0;
while(l1.size()!=0&&l2.size()!=0){
int x=(int)l1.removeFirst()+(int)l2.removeFirst()+carry;
int r=(x)%10;
carry=(x)/10;
while(l1.size()!=0){
int x=(int)l1.removeFirst()+carry;
int r=(x)%10;
carry=(x)/10;
while(l2.size()!=0){
int x=(int)l2.removeFirst()+carry;
int r=(x)%10;
carry=(x)/10;
if(carry==1)
return res;}

• 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;

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?