**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);
if (head == NULL) {
curr = newnode;
head = 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;
}
}
return head;
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
public LLNode solve(LLNode l0, LLNode l1) {
LLNode head = new LLNode();
LLNode tail = head;
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;
return head.next;
}
}
```

` ````
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.
head, tail = None, None
while l0 is not None and l1 is not None:
if l0.val == l1.val:
if head is None:
head = LLNode(l0.val)
tail = head
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
return head
```

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