### Remove nth Node from Last - Leet Code Solution

Problem Statement Given a linked list, remove the n-th node from the end of list…

July 23, 2019

You are given two non-empty linked lists representing two non-negative integers. 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.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

```
Example:
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.
```

```
private static ListNode addTwoNumbersHelper(ListNode l1, ListNode l2, int carry) {
if (l1 == null && l2 == null && carry == 0) {
return null;
}
int sum = carry;
if (l1 != null) { sum += l1.val; }
if (l2 != null) { sum += l2.val; }
if (sum > 9) {
sum = sum - 10;
carry = 1;
}
else {
carry = 0;
}
ListNode newNode = new ListNode(sum);
newNode.next = addTwoNumbersHelper(l1 != null ? l1.next : null, l2 != null ? l2.next : null, carry);
return newNode;
}
public static ListNode addTwoNumbers(ListNode l1, ListNode l2) {
return addTwoNumbersHelper(l1, l2, 0);
}
```

This problem is rather simple. Since, the number is represented as link list in reverse order. i.e. the first node represent the last digit of a number. So, you can just started adding number as you traverse the two link list.

- Start with iterating both link list.
- take sum of two nodes, and carry if previously set.
- If the sum is more than single digit, set the carry.
- Now, new sum is our new node.
- Since, it will be last digit in the result number, and will be first node in link list.
- Its next pointer will be the one which will come out in next round of sum.
- So, we simply set its next and call same method in recursive manner. Which will again goes in recursive mode untill all link list is exhausted.
- And, we get our link list as sum.

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