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.
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:
listA is in the m.listB is in the n.1 <= m, n <= 3 * 1041 <= Node.val <= 1050 <= skipA < m0 <= skipB < nintersectVal 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?The core challenge of this problem is to identify the node at which two singly linked lists intersect. This problem is significant in various applications such as finding common elements in data streams, merging paths in routing algorithms, and more. A common pitfall is assuming that the lists intersect at the same position from the start, which is not necessarily true.
To solve this problem, we need to consider the following steps:
A naive solution would involve using nested loops to compare each node of the first list with each node of the second list. This approach has a time complexity of O(m * n), which is not optimal for large lists.
An optimized solution involves using two pointers to traverse the lists. This approach ensures that both pointers traverse the same number of nodes, making it possible to find the intersection in O(m + n) time complexity.
Here is a step-by-step breakdown of the optimized algorithm:
pA and pB, to the heads of headA and headB respectively.pA reaches the end of headA, redirect it to the head of headB. Similarly, when pB reaches the end of headB, redirect it to the head of headA.pA and pB meet at the intersection node or both reach the end (null)./**
* Definition for singly-linked list.
* function ListNode(val) {
* this.val = val;
* this.next = null;
* }
*/
/**
* @param {ListNode} headA
* @param {ListNode} headB
* @return {ListNode}
*/
var getIntersectionNode = function(headA, headB) {
// Initialize two pointers
let pA = headA;
let pB = headB;
// Traverse the lists
while (pA !== pB) {
// If pA reaches the end of list A, redirect it to the head of list B
pA = pA === null ? headB : pA.next;
// If pB reaches the end of list B, redirect it to the head of list A
pB = pB === null ? headA : pB.next;
}
// Either both pointers meet at the intersection node or both are null
return pA;
};
The time complexity of this approach is O(m + n), where m and n are the lengths of the two linked lists. The space complexity is O(1) as we are using only two pointers and no additional data structures.
Consider the following edge cases:
Our algorithm handles these cases effectively by ensuring that the pointers traverse the same number of nodes.
To test the solution comprehensively, consider the following test cases:
// Test Case 1: Intersecting lists
let intersectVal = 8;
let listA = [4,1,8,4,5];
let listB = [5,6,1,8,4,5];
let skipA = 2;
let skipB = 3;
// Expected Output: Intersected at '8'
// Test Case 2: Intersecting lists
intersectVal = 2;
listA = [1,9,1,2,4];
listB = [3,2,4];
skipA = 3;
skipB = 1;
// Expected Output: Intersected at '2'
// Test Case 3: Non-intersecting lists
intersectVal = 0;
listA = [2,6,4];
listB = [1,5];
skipA = 3;
skipB = 2;
// Expected Output: No intersection
When approaching such problems, consider the following tips:
In this blog post, we discussed how to find the intersection of two linked lists in O(m + n) time complexity using JavaScript. We covered the problem definition, approach, algorithm, code implementation, complexity analysis, edge cases, and testing. Understanding and solving such problems is crucial for improving your problem-solving skills and preparing for technical interviews.
For further reading and practice, consider the following resources: