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.


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


l0 = [1, 3, 7]

l1 = [2, 3, 7, 9]


[3, 7]

Example 2


l0 = [1, 2, 3]

l1 = [4, 5, 6]



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;
                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) {
       = l0;
                tail =;

                l0 =;
                l1 =;
            } else if (l0.val > l1.val)
                l1 =;
                l0 =;
        } = 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.
        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
           = LLNode(l0.val)
                    tail =
                l0, l1 =,
            elif l0.val < l1.val:
                l0 =
            elif l1.val < l0.val:
                l1 =
        return head

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