Last updated on October 10th, 2024 at 12:54 am

Here, We see Intersection of Two Linked Lists LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc., with different approaches.

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Intersection of Two Linked Lists LeetCode Solution

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: (fig-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: (fig-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: (fig-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.

1. Intersection of Two Linked Lists Leetcode Solution C++

class Solution {
public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        ListNode *ptr1 = headA;
		ListNode *ptr2 = headB;
		while(ptr1 != ptr2){
			if(ptr1 == NULL){
				ptr1 = headB;
			}
			else{
				ptr1 = ptr1 -> next;
			}
			if(ptr2 == NULL){
				ptr2 = headA;
			}
			else{
				ptr2 = ptr2 -> next;
			}
		}
		return ptr1;
    }
};

1.1 Explanation

• Variables: ptr1 and ptr2 are initialized to headA and headB respectively, representing pointers to traverse the two linked lists.
• Traversal: The while loop continues until ptr1 equals ptr2, which indicates they have found the intersection node.
• Pointer Manipulation:
  • If ptr1 reaches the end (NULL), it continues from headB.
  • If ptr2 reaches the end (NULL), it continues from headA.
• Return: Once ptr1 equals ptr2, they both point to the intersection node, and it is returned.

1.2 Time Complexity

• O(m + n), where m and n are the lengths of headA and headB respectively.

1.3 Space Complexity

• O(1), since only a constant amount of extra space is used for pointers.

2. Intersection of Two Linked Lists Leetcode Solution Java

public class Solution {
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        ListNode a = headA, b = headB;
        while (a != b) {
            a = a == null ? headB : a.next;
            b = b == null ? headA : b.next;
        }
        return a;
    }
}

2.1 Explanation

• Variables: a and b are initialized to headA and headB respectively.
• Traversal: The while loop continues until a equals b, indicating they have found the intersection node.
• Pointer Manipulation:
  • If a reaches the end (null), it continues from headB.
  • If b reaches the end (null), it continues from headA.
• Return: Once a equals b, they both point to the intersection node, and it is returned.

2.2 Time Complexity

• O(m + n), where m and n are the lengths of headA and headB respectively. Each pointer might traverse both lists once.

2.3 Space Complexity

• O(1), since only a constant amount of extra space is used for pointers.

3. Intersection of Two Linked Lists Leetcode Solution JavaScript

var getIntersectionNode = function(headA, headB) {
    if (!headA || !headB) {
        return null;
    }
    let curA = headA;
    let curB = headB;
    while (curA !== curB) {
        
        if (curA.next) {
            curA = curA.next;
        } else {
        
            if (!curB.next) {
                curA = null;
                curB = null;
                break;
            }
            curA = headB;
        }
        
        if (curB.next) {
            curB = curB.next 
        } else {
            curB = headA;
        }
    }
    return curB;
};

3.1 Explanation

• Edge Case: If either list is empty (!headA or !headB), return null.
• Variables: curA and curB are initialized to headA and headB respectively.
• Traversal: The while loop continues until curA equals curB, indicating they have found the intersection node.
• Pointer Manipulation:
  • If curA.next exists, move curA to its next node; otherwise, redirect curA to headB.
  • If curB.next exists, move curB to its next node; otherwise, redirect curB to headA.
• Return: Once curA equals curB, they both point to the intersection node, and it is returned.

3.2 Time Complexity

• O(m + n), where m and n are the lengths of headA and headB respectively. Each pointer might traverse both lists once.

3.3 Space Complexity

• O(1), since only a constant amount of extra space is used for pointers.

4. Intersection of Two Linked Lists Leetcode Solution Python

class Solution(object):
    def getIntersectionNode(self, headA, headB):
        curA,curB = headA,headB
        lenA,lenB = 0,0
        while curA is not None:
            lenA += 1
            curA = curA.next
        while curB is not None:
            lenB += 1
            curB = curB.next
        curA,curB = headA,headB
        if lenA > lenB:
            for i in range(lenA-lenB):
                curA = curA.next
        elif lenB > lenA:
            for i in range(lenB-lenA):
                curB = curB.next
        while curB != curA:
            curB = curB.next
            curA = curA.next
        return curA

4.1 Explanation

• Variables: curA and curB are initialized to headA and headB respectively, to traverse the lists.
• Length Calculation: Count the lengths of both lists using lenA and lenB.
• Adjustment:
  • Adjust curA or curB to start from the same relative position based on the difference in lengths.
• Traversal: Traverse both lists simultaneously until curA equals curB, which indicates the intersection node.
• Return: Return curA (or curB since they are equal).

4.2 Time Complexity

• O(m + n), where m and n are the lengths of headA and headB respectively.

4.3 Space Complexity

• O(1), since only a constant amount of extra space is used for pointers and variables.