Table of Contents

## Problem Statement

Intersection of Two Linked Lists LeetCode Solution – We are given the heads of two strongly linked-lists headA and headB. It is also given that the two linked lists may intersect at some point. We are asked to return the node at which they intersect or null if they have no intersection.

## Examples & Explanation

**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.

**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.

## Approach

The idea is very intuitive, we can simply find the difference between the lengths of given linked lists and change the root or the head node of the longer list to this difference. Keep iterating the nodes till an intersection is encountered. If one of the nodes becomes NULL or nullptr, then there exists no intersection, return NULL, otherwise return node.

## Code

### C++ code for Intersection of Two Linked Lists

/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) { // declaring integers to store the size of the linked lists int a=0, b=0; // to count the size of the linked list iterate through each node & increment size ListNode *currA = headA, *currB = headB; while(currA != NULL){ currA = currA -> next; a++; } while(currB != NULL) { currB = currB->next; b++; } int k = abs(b-a); if ( b > a ) { //if the size of B > A, then b becomes b->next till the difference a-b becomes zero while(k--) { headB = headB -> next; } } else { //if the size of B < A, then a becomes a->next till the difference a-b becomes zero while(k--) { headA = headA -> next; } } // check if the headA and headB are equal, if not then move to next pointers of headA and headB while(headA != headB && headA->next != NULL && headB->next != NULL) { headA = headA -> next; headB = headB -> next; } // if one of the list becomes NULL and still no node is found common then return NULL if(headA != headB) return NULL; // if above statement is not true, then return the common node either headA or headB return headA; } };

### Java code for Intersection of Two Linked Lists

/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { * val = x; * next = null; * } * } */ public class Solution { public ListNode getIntersectionNode(ListNode headA, ListNode headB) { int a=0, b=0; ListNode currA = headA, currB = headB; while(currA != null){ currA = currA.next; a++; } while(currB != null) { currB = currB.next; b++; } int k = Math.abs(b-a); if ( b > a ) { while(k>0) { headB = headB.next; k--; } } else { while(k>0) { headA = headA.next; k--; } } while(headA != headB && headA.next != null && headB.next != null) { headA = headA.next; headB = headB.next; } if(headA != headB) return null; // if above statement is not true, then return the common node either headA or headB return headA; } }

**Complexity Analysis for Intersection of Two Linked Lists LeetCode Solution**

Let N be the length of list A and M be the length of list B.

- Time complexity : O(N + M)
- In the worst-case i.e. intersection at the last node, each list will be traversed twice taking O(N+M) time. Note that after the longer list has traversed ‘k’ times, or we can say when the lists are lined up, this algorithm traverses each list node only once. This is because the pointers firstly go down each list so that they can be “lined up” and then in the second iteration, the intersection node is searched for.

- Space complexity: O(1)
- We aren’t allocating any additional data structures, so the amount of extra space used does not grow with the size of the input.