# Intersection of Two Linked Lists

Write a program to find the node at which the intersection of two singly linked lists begins.

For example, the following two linked lists:

``````A:          a1 → a2
↘
c1 → c2 → c3
↗
B:     b1 → b2 → b3``````

begin to intersect at node c1.

Notes:

• If the two linked lists have no intersection at all, return `null`

• The linked lists must retain their original structure after the function returns.

• You may assume there are no cycles anywhere in the entire linked structure.

• Your code should preferably run in O(n) time and use only O(1) memory.

## Analysis

This problem at first doesn't appear to be hard, and it's easy to come up with Brute Force or Hash Table approaches.

But what's really fascinating is coming up with an idea of combining two linked list in order to make use of Two Pointers technique.

### Approach 1: Brute Force

For each node `ai` in list A, traverse the entire list B and check if any node in list B coincides with `ai`.

Complexity Analysis

• Time complexity : O(mn).

• Space complexity : O(1).

### Approach 2: Hash Table

Traverse list A and store the address / reference to each node in a hash set. Then check every node `bi` in list B: if `bi` appears in the hash set, then `bi`is the intersection node.

Complexity Analysis

• Time complexity :O(m+n).

• Space complexity :O(m) or O(n).

### Approach 3: Two Pointers

Maintain two pointers pA and pB initialized at the head of A and B, respectively. Then let them both traverse through the lists, one node at a time.

When pA reaches the end of a list, then redirect it to the head of B (yes, B, that's right.); similarly when pB reaches the end of a list, redirect it the head of A.

If at any point pA meets pB, then pA/pB is the intersection node.

To see why the above trick would work, consider the following two lists: A = {1,3,5,7,9,11} and B = {2,4,9,11}, which are intersected at node '9'. Since B.length (=4) < A.length (=6), pB would reach the end of the merged list first, because pB traverses exactly 2 nodes less than pA does. By redirecting pB to head A, and pA to head B, we now ask pB to travel exactly 2 more nodes than pA would. So in the second iteration, they are guaranteed to reach the intersection node at the same time.

If two lists have intersection, then their last nodes must be the same one. So when pA/pB reaches the end of a list, record the last element of A/B respectively. If the two last elements are not the same one, then the two lists have no intersections.

Complexity Analysis

• Time complexity : O(m+n).

• Space complexity : O(1).

Here's another nice visualization of this solution:

## Solution

Two Pointers - O(n) time, O(1) space

``````/**
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) {
*         val = x;
*         next = null;
*     }
* }
*/
public class Solution {