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Intersection of Two Linked Lists - Solution & Explanation

EasyHash TableLinked ListTwo Pointers21 min readAsked at: Amazon, Microsoft, Apple +10
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Problem Statement

Given the heads of two singly linked-lists headA and headB, return the node at which the two lists intersect. If the two linked lists have no intersection at all, return null.

For example, the following two linked lists begin to intersect at node c1:

The test cases are generated such that there are no cycles anywhere in the entire linked structure.

Note that the linked lists must retain their original structure after the function returns.

Custom Judge:

The inputs to the judge are given as follows (your program is not given these inputs):

  • intersectVal - The value of the node where the intersection occurs. This is 0 if there is no intersected node.
  • listA - The first linked list.
  • listB - The second linked list.
  • skipA - The number of nodes to skip ahead in listA (starting from the head) to get to the intersected node.
  • skipB - The number of nodes to skip ahead in listB (starting from the head) to get to the intersected node.

The judge will then create the linked structure based on these inputs and pass the two heads, headA and headB to your program. If you correctly return the intersected node, then your solution will be accepted.

 

Example 1:

Input: intersectVal = 8, listA = [4,1,8,4,5], listB = [5,6,1,8,4,5], skipA = 2, skipB = 3
Output: Intersected at '8'
Explanation: The intersected node's value is 8 (note that this must not be 0 if the two lists intersect).
From the head of A, it reads as [4,1,8,4,5]. From the head of B, it reads as [5,6,1,8,4,5]. There are 2 nodes before the intersected node in A; There are 3 nodes before the intersected node in B.
- Note that the intersected node's value is not 1 because the nodes with value 1 in A and B (2nd node in A and 3rd node in B) are different node references. In other words, they point to two different locations in memory, while the nodes with value 8 in A and B (3rd node in A and 4th node in B) point to the same location in memory.

Example 2:

Input: intersectVal = 2, listA = [1,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1
Output: Intersected at '2'
Explanation: The intersected node's value is 2 (note that this must not be 0 if the two lists intersect).
From the head of A, it reads as [1,9,1,2,4]. From the head of B, it reads as [3,2,4]. There are 3 nodes before the intersected node in A; There are 1 node before the intersected node in B.

Example 3:

Input: intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2
Output: No intersection
Explanation: From the head of A, it reads as [2,6,4]. From the head of B, it reads as [1,5]. Since the two lists do not intersect, intersectVal must be 0, while skipA and skipB can be arbitrary values.
Explanation: The two lists do not intersect, so return null.

 

Constraints:

  • The number of nodes of listA is in the m.
  • The number of nodes of listB is in the n.
  • 1 <= m, n <= 3 * 104
  • 1 <= Node.val <= 105
  • 0 <= skipA <= m
  • 0 <= skipB <= n
  • intersectVal is 0 if listA and listB do not intersect.
  • intersectVal == listA[skipA] == listB[skipB] if listA and listB intersect.

 

Follow up: Could you write a solution that runs in O(m + n) time and use only O(1) memory?

Approach Overview

Problem Overview: You are given the heads of two singly linked lists. The task is to return the node where the two lists intersect. After the intersection point, both lists share the exact same nodes in memory. If the lists never meet, return null. The challenge is detecting the shared node without modifying the lists and while keeping time and space efficient.

Approach 1: Using Length Difference (O(m+n) time, O(1) space)

First compute the length of both lists by iterating through them once. If one list is longer, advance its pointer by the length difference so both pointers are equally far from the tail. Now move both pointers forward together one node at a time. The first node where the pointers are equal is the intersection node. This approach relies on the property that once the lists align at the same remaining distance, the shared portion will be reached simultaneously. This method is straightforward if you are comfortable with linked list traversal.

Approach 2: Two Pointer Technique (O(m+n) time, O(1) space)

Use two pointers starting at the heads of the two lists. When a pointer reaches the end of its list, redirect it to the head of the other list. After at most two passes, both pointers travel the same total distance. If an intersection exists, the pointers meet exactly at the intersection node. If not, they both reach null at the same time. The key insight is distance equalization: each pointer effectively traverses lenA + lenB. This elegant trick is a classic application of the two pointers pattern and avoids explicitly computing lengths.

Recommended for interviews: The two pointer switching technique is what most interviewers expect. It demonstrates a deeper understanding of pointer traversal and avoids extra preprocessing. The length difference method still shows solid reasoning and is often the first solution candidates derive. Both run in O(m+n) time with O(1) extra space, which is optimal for this linked list problem.

Approach 1: Approach 1: Using Length Difference

First, traverse both linked lists to determine their lengths. Calculate the difference in lengths and advance the pointer of the longer list by the length difference. Then move both pointers in tandem to find the intersection node.

This solution first determines the lengths of both lists. It advances the pointer for the longer list by the difference in lengths. Then, it compares nodes in both lists step by step to find the intersection.

Code

CC++JavaPythonC#JavaScript

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Complexity

Time Complexity: O(m + n).
Space Complexity: O(1).

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Approach 2: Approach 2: Two Pointer Technique

Use two pointers, each starting at the head of one list. Traverse the list until a pointer reaches null, then start traversing the other list from the beginning. Repeat until the two pointers meet at the intersection node.

This implementation uses two pointers initialized to the heads of the two lists. Each pointer traverses its list and switches to the other list once it reaches the end. They meet at the intersection node or at null if there is none.

Code

CC++JavaPythonC#JavaScript

C

C++

Java

Python

C#

JavaScript

Complexity

Time Complexity: O(m + n).
Space Complexity: O(1).

Try this approach in the editor →

Approach 3: Two Pointers

We use two pointers a and b to point to the heads of the two linked lists headA and headB, respectively.

Traverse the linked lists simultaneously. When a reaches the end of headA, redirect it to the head of headB. Similarly, when b reaches the end of headB, redirect it to the head of headA.

If the two pointers meet, the node they point to is the first common node. If they do not meet, it means the two linked lists have no common nodes, and both pointers will point to null. Return either pointer.

The time complexity is O(m + n), where m and n are the lengths of the linked lists headA and headB, respectively. The space complexity is O(1).

Code

PythonJavaC++GoTypeScriptJavaScriptSwift

Python

Java

C++

Go

TypeScript

JavaScript

Swift

Try this approach in the editor →

Complexity Comparison

ApproachComplexity
Approach 1: Using Length Difference

Time Complexity: O(m + n).
Space Complexity: O(1).

Approach 2: Two Pointer Technique

Time Complexity: O(m + n).
Space Complexity: O(1).

Two Pointers—

Detailed Complexity Analysis

ApproachTimeSpaceWhen to Use
Length Difference AlignmentO(m+n)O(1)When you want a straightforward approach by aligning list lengths before traversal
Two Pointer Switching TechniqueO(m+n)O(1)Best general solution; commonly expected in coding interviews

Video Solution

Intersection of Two Linked Lists - Leetcode 160 - Python • NeetCode • 105,351 views views

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Frequently Asked Questions

Is Intersection of Two Linked Lists easy or hard?
Intersection of Two Linked Lists is categorized as an Easy problem on LeetCode with an acceptance rate around 63%. The difficulty mainly comes from discovering the pointer switching trick that equalizes traversal distance without computing list lengths.
Intersection of Two Linked Lists Python/Java solution
Python and Java implementations typically follow the two pointer approach. Maintain two references starting at each list head. When a pointer reaches the end, redirect it to the other list. Continue until both pointers meet or become null. The logic is concise and works in O(m+n) time with O(1) space.
How to solve Intersection of Two Linked Lists in O(n)?
Use the two pointer technique. Initialize pointers at the heads of both lists. When one pointer reaches the end, move it to the other list's head. After at most two passes, both pointers align and either meet at the intersection node or reach null simultaneously. This ensures O(m+n) time and constant space.
What is the best approach for Intersection of Two Linked Lists?
The two pointer switching technique is considered the best approach. Start one pointer at each list head. When a pointer reaches the end, redirect it to the head of the other list. Both pointers traverse equal total distance (lenA + lenB), so they meet at the intersection node in O(m+n) time with O(1) space.
Is Intersection of Two Linked Lists asked at Google/Amazon/Meta?
Intersection of Two Linked Lists is a common linked list interview question asked by companies like Amazon, Google, Microsoft, and Meta. It tests pointer manipulation, linked list traversal, and the ability to optimize for O(1) space.
What data structure is used in Intersection of Two Linked Lists?
The problem is based on singly linked lists and pointer traversal. Some alternative solutions use a hash table to store visited nodes, but the optimal approaches rely only on pointer movement and linked list properties without additional data structures.
What is the time complexity of Intersection of Two Linked Lists?
The optimal algorithms run in O(m+n) time where m and n are the lengths of the two linked lists. Each pointer traverses both lists at most once. Space complexity is O(1) because only a few pointer variables are used without extra data structures.

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