Intersection of Two Linked Lists solution and explanation
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The problem
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 is0if there is no intersected node.listA- The first linked list.listB- The second linked list.skipA- The number of nodes to skip ahead inlistA(starting from the head) to get to the intersected node.skipB- The number of nodes to skip ahead inlistB(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
listAis in them. - The number of nodes of
listBis in then. 1 <= m, n <= 3 * 1041 <= Node.val <= 1050 <= skipA <= m0 <= skipB <= nintersectValis0iflistAandlistBdo not intersect.intersectVal == listA[skipA] == listB[skipB]iflistAandlistBintersect.
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.
class ListNode:
def __init__(self, x):
self.val = x
self.next = None
class Solution:
def getLength(self, head):
length = 0
while head:
length += 1
head = head.next
return length
def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode:
lenA = self.getLength(headA)
lenB = self.getLength(headB)
if lenA > lenB:
for _ in range(lenA - lenB):
headA = headA.next
else:
for _ in range(lenB - lenA):
headB = headB.next
while headA != headB:
headA = headA.next
headB = headB.next
return headApublic class Solution {
public class ListNode {
int val;
ListNode next;
ListNode(int x) {
val = x;
next = null;
}
}
private int getLength(ListNode head) {
int length = 0;
while (head != null) {
length++;
head = head.next;
}
return length;
}
public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
int lenA = getLength(headA);
int lenB = getLength(headB);
if (lenA > lenB) {
for (int i = 0; i < lenA - lenB; i++) headA = headA.next;
} else {
for (int i = 0; i < lenB - lenA; i++) headB = headB.next;
}
while (headA != headB) {
headA = headA.next;
headB = headB.next;
}
return headA;
}
}#include <iostream>
using namespace std;
struct ListNode {
int val;
ListNode* next;
ListNode(int x) : val(x), next(NULL) {}
};
int getLength(ListNode* head) {
int length = 0;
while (head) {
length++;
head = head->next;
}
return length;
}
ListNode* getIntersectionNode(ListNode* headA, ListNode* headB) {
int lenA = getLength(headA);
int lenB = getLength(headB);
if (lenA > lenB) {
for (int i = 0; i < lenA - lenB; i++) headA = headA->next;
} else {
for (int i = 0; i < lenB - lenA; i++) headB = headB->next;
}
while (headA != headB) {
headA = headA->next;
headB = headB->next;
}
return headA;
}function ListNode(val) {
this.val = val;
this.next = null;
}
function getLength(head) {
let length = 0;
while (head) {
length++;
head = head.next;
}
return length;
}
function getIntersectionNode(headA, headB) {
let lenA = getLength(headA);
let lenB = getLength(headB);
if (lenA > lenB) {
for (let i = 0; i < lenA - lenB; i++) headA = headA.next;
} else {
for (let i = 0; i < lenB - lenA; i++) headB = headB.next;
}
while (headA !== headB) {
headA = headA.next;
headB = headB.next;
}
return headA;
}#include <stdio.h>
#include <stdlib.h>
struct ListNode {
int val;
struct ListNode *next;
};
int getLength(struct ListNode *head) {
int length = 0;
while (head) {
length++;
head = head->next;
}
return length;
}
struct ListNode *getIntersectionNode(struct ListNode *headA, struct ListNode *headB) {
int lenA = getLength(headA);
int lenB = getLength(headB);
if (lenA > lenB) {
for (int i = 0; i < lenA - lenB; i++)
headA = headA->next;
} else {
for (int i = 0; i < lenB - lenA; i++)
headB = headB->next;
}
while (headA != headB) {
headA = headA->next;
headB = headB->next;
}
return headA;
}public class Solution {
public class ListNode {
public int val;
public ListNode next;
public ListNode(int x) { val = x; next = null; }
}
private int GetLength(ListNode head) {
int length = 0;
while (head != null) {
length++;
head = head.next;
}
return length;
}
public ListNode GetIntersectionNode(ListNode headA, ListNode headB) {
int lenA = GetLength(headA);
int lenB = GetLength(headB);
if (lenA > lenB) {
for (int i = 0; i < lenA - lenB; i++) headA = headA.next;
} else {
for (int i = 0; i < lenB - lenA; i++) headB = headB.next;
}
while (headA != headB) {
headA = headA.next;
headB = headB.next;
}
return headA;
}
}The Python solution calculates the length of each list. If one is longer, it advances that pointer by the difference. Both pointers are then compared step by step until the intersection is found.
Time Complexity: O(m + n).
Space Complexity: O(1).
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.
class ListNode:
def __init__(self, x):
self.val = x
self.next = None
class Solution:
def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode:
if not headA or not headB:
return None
a, b = headA, headB
while a != b:
a = a.next if a else headB
b = b.next if b else headA
return apublic class Solution {
public class ListNode {
int val;
ListNode next;
ListNode(int x) {
val = x;
next = null;
}
}
public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
if (headA == null || headB == null) return null;
ListNode a = headA;
ListNode b = headB;
while (a != b) {
a = (a == null) ? headB : a.next;
b = (b == null) ? headA : b.next;
}
return a;
}
}#include <iostream>
using namespace std;
struct ListNode {
int val;
ListNode* next;
ListNode(int x) : val(x), next(NULL) {}
};
ListNode* getIntersectionNode(ListNode* headA, ListNode* headB) {
if (!headA || !headB) return NULL;
ListNode* a = headA;
ListNode* b = headB;
while (a != b) {
a = a ? a->next : headB;
b = b ? b->next : headA;
}
return a;
}function ListNode(val) {
this.val = val;
this.next = null;
}
function getIntersectionNode(headA, headB) {
if (!headA || !headB) return null;
let a = headA;
let b = headB;
while (a !== b) {
a = a ? a.next : headB;
b = b ? b.next : headA;
}
return a;
}#include <stdio.h>
#include <stdlib.h>
struct ListNode {
int val;
struct ListNode *next;
};
struct ListNode *getIntersectionNode(struct ListNode *headA, struct ListNode *headB) {
if (!headA || !headB) return NULL;
struct ListNode *a = headA;
struct ListNode *b = headB;
while (a != b) {
a = a ? a->next : headB;
b = b ? b->next : headA;
}
return a;
}public class Solution {
public class ListNode {
public int val;
public ListNode next;
public ListNode(int x) { val = x; next = null; }
}
public ListNode GetIntersectionNode(ListNode headA, ListNode headB) {
if (headA == null || headB == null) return null;
ListNode a = headA;
ListNode b = headB;
while (a != b) {
a = (a == null) ? headB : a.next;
b = (b == null) ? headA : b.next;
}
return a;
}
}The Python solution follows the two pointer method where each pointer traverses its list, and when reaching the end, starts from the beginning of the other list. This continues until they meet at an intersection node or at the end as null.
Time Complexity: O(m + n).
Space Complexity: O(1).
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).
/**
* Definition for singly-linked list.
* class ListNode {
* val: number
* next: ListNode | null
* constructor(val?: number, next?: ListNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.next = (next===undefined ? null : next)
* }
* }
*/
function getIntersectionNode(headA: ListNode | null, headB: ListNode | null): ListNode | null {
let [a, b] = [headA, headB];
while (a !== b) {
a = a ? a.next : headB;
b = b ? b.next : headA;
}
return a;
}/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func getIntersectionNode(headA, headB *ListNode) *ListNode {
a, b := headA, headB
for a != b {
if a == nil {
a = headB
} else {
a = a.Next
}
if b == nil {
b = headA
} else {
b = b.Next
}
}
return a
}/**
* Definition for singly-linked list.
* public class ListNode {
* public var val: Int
* public var next: ListNode?
* public init(_ val: Int) {
* self.val = val
* self.next = nil
* }
* }
*/
class Solution {
func getIntersectionNode(_ headA: ListNode?, _ headB: ListNode?) -> ListNode? {
var a = headA
var b = headB
while a !== b {
a = a == nil ? headB : a?.next
b = b == nil ? headA : b?.next
}
return a
}
}Complexity comparison
| Approach | Complexity |
|---|---|
| Approach 1: Using Length Difference | Time Complexity: O(m + n). |
| Approach 2: Two Pointer Technique | Time Complexity: O(m + n). |
| Two Pointers | – |
Complexity analysis
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Length Difference Alignment | O(m+n) | O(1) | When you want a straightforward approach by aligning list lengths before traversal |
| Two Pointer Switching Technique | O(m+n) | O(1) | Best general solution; commonly expected in coding interviews |
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.