**Add Linked Lists - Amazon Top Interview Questions**

### Problem Statement :

Given a singly linked list l0 and another linked list l1, each representing a number with least significant digits first, return the summed linked list. Note: Each value in the linked list is guaranteed to be between 0 and 9. Constraints n ≤ 100,000 where n is the number of nodes in l0 m ≤ 100,000 where m is the number of nodes in l1 Example 1 Input l0 = [6, 4] l1 = [4, 7] Output [0, 2, 1] Explanation This is 46 + 74 = 120

### Solution :

` ````
Solution in C++ :
/**
* class LLNode {
* public:
* int val;
* LLNode *next;
* };
*/
LLNode* solve(LLNode* l0, LLNode* l1) {
int carry = 0;
LLNode *head = new LLNode(0), *res = head;
// now normal addition :D
// add the numbers, then send over the carry
while (l0 or l1 or carry) {
res->next = new LLNode();
res = res->next;
int a, b;
if (!l0)
a = 0;
else
a = l0->val, l0 = l0->next;
if (!l1)
b = 0;
else
b = l1->val, l1 = l1->next;
// now just add these two to each other!
int digit = (a + b + carry) % 10;
carry = (a + b + carry) / 10;
res->val = digit;
// cout << " added " << digit << " to the list " << endl;
}
return head->next;
}
```

` ````
Solution in Python :
# class LLNode:
# def __init__(self, val, next=None):
# self.val = val
# self.next = next
class Solution:
def solve(self, l0, l1):
def finish_remaining(l, c):
s, head, tail = 0, None, None
while l is not None: # Need to do extra addition only if there is a leftover carry
entry = l.val + c
s = entry % 10
c = entry // 10
l.val = s
if head is None:
head = tail = l
else:
tail.next = l
tail = l
prev, l = l, l.next
if c == 1: # Try l =[9,9] and c=1 what should happen?
prev.next = LLNode(1)
return head
if l0 is None and l1 is not None:
return l1
elif l1 is None and l0 is not None:
return l0
elif l0 is None and l1 is None:
return None
s, c, prev, head = 0, 0, None, l0
while l0 is not None and l1 is not None:
entry = l0.val + l1.val + c
s, c = entry % 10, entry // 10
l0.val = s
prev, l0, l1 = l0, l0.next, l1.next
prev.next = None
if l0 is None and l1 is not None:
prev.next = finish_remaining(l1, c)
elif l1 is None and l0 is not None:
prev.next = finish_remaining(l0, c)
elif c == 1: # l1 and l0 are both None But there is a leftover carry. Try for Example 1
prev.next = LLNode(1)
return head
```

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