# Linked List Intersection - Amazon Top Interview Questions

### Problem Statement :

```Given two sorted linked lists l0, and l1, return a new sorted linked list containing the intersection of the two lists.

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 = [1, 3, 7]

l1 = [2, 3, 7, 9]

Output

[3, 7]

Example 2

Input

l0 = [1, 2, 3]

l1 = [4, 5, 6]

Output

None```

### Solution :

```                        ```Solution in C++ :

LLNode* solve(LLNode* l0, LLNode* l1) {
if (l0 == NULL or l1 == NULL) return NULL;

LLNode *curr = NULL, *head = NULL;
LLNode* ptr1 = l0;
LLNode* ptr2 = l1;

while (ptr2 and ptr1) {
if (ptr2->val == ptr1->val) {
LLNode* newnode = new LLNode(ptr1->val);
curr = newnode;
} else {
curr->next = newnode;
curr = curr->next;
}

ptr1 = ptr1->next;
ptr2 = ptr2->next;
}

else {
if (ptr1->val > ptr2->val)
ptr2 = ptr2->next;
else
ptr1 = ptr1->next;
}
}

}```
```

```                        ```Solution in Java :

import java.util.*;

class Solution {
public LLNode solve(LLNode l0, LLNode l1) {

while (l0 != null && l1 != null) {
if (l0.val == l1.val) {
tail.next = l0;
tail = tail.next;

l0 = l0.next;
l1 = l1.next;
} else if (l0.val > l1.val)
l1 = l1.next;
else
l0 = l0.next;
}

tail.next = null;

}
}```
```

```                        ```Solution in Python :

class Solution:
def solve(self, l0, l1):
# Remember the lists are sorted
# Also remember that intersection of two linked lists does not mean the lists actually intersect
# like the case where l0 : 3--4---
#                                 |
#                      l1   1--2--5--6--7
# like l1 is not a part of l0 or vice versa
# This is just plain intersection where we have to find elements that are common in both.
while l0 is not None and l1 is not None:
if l0.val == l1.val:
else:
tail.next = LLNode(l0.val)
tail = tail.next
l0, l1 = l0.next, l1.next
elif l0.val < l1.val:
l0 = l0.next
elif l1.val < l0.val:
l1 = l1.next
```

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