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 (2^nd node in A and 3^rd 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 (3^rd node in A and 4^th node in B) point to the same location in memory.

Intuition

The problem states that we need to find the intersection of two linked lists without changing their original structure.

Approach 1

1. Find the length of both the linked lists.
2. Traverse the bigger linked list until the remaining nodes count becomes equal to the smaller one’s.
3. Traverse both the heads together. If both of them are same then the intersecting node is found.

Complexity

• Time complexity: O(n) where n>m
• Space complexity: O(1)

class Solution {
public:
    int length(ListNode *head){
        int len = 0;
        while(head){
            len++;
            head = head->next;
        }
        return len;
    }

    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        if(!headA || !headB) return NULL;

        //step1
        int lenA = length(headA), lenB = length(headB);

        //step2
        if(lenA>lenB){
            while(lenA>lenB){
                headA = headA->next;
                lenA--;
            }
        }
        else if(lenA<lenB){
            while(lenA<lenB){
                headB = headB->next;
                lenB--;
            }
        }
        
        //step 3
        while(headA && headB){
            if(headA==headB) return headA;
            headA = headA->next;
            headB = headB->next;
        }
        return NULL;
    }
};

Approach 2

Prerequisite : https://leetcode.com/problems/linked-list-cycle-ii/

In this approach, we can simply convert this problem into a loop problem.

1. Find the tail.
2. Connect the tail with any of the head which creates a loop.
3. Using the other head, find intersection point of the loop.
4. Undo the loop, by setting tail->next = NULL
5. Return the intersection node.

Complexity

• Time complexity: O(n+m)
• Space complexity: O(1)

ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
    //getting the tail
    ListNode* tail = headA;
    while(tail->next){
        tail = tail->next;
    }

    //creating a loop
    tail->next = headA;

    //detecting and finding the intersection
    ListNode *fast = headB, *slow = headB;

    while(fast && fast->next){
        slow = slow->next;
        fast = fast->next->next;

        if(slow==fast) {
            slow = headB;
            while(slow!=fast){
                slow = slow->next;
                fast = fast->next;
            }
            //undoing the loop
            tail->next = NULL;
            return slow;
        };
    }
    tail->next = NULL;
    return NULL;
}