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 (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:
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?Problem Overview: Two singly linked lists may merge at some node and share the rest of their nodes. The task is to return the exact node where both lists intersect. If they never intersect, return null. The key constraint: nodes are shared by reference, not by value.
Approach 1: Using Length Difference (O(n + m) time, O(1) space)
This method first measures the lengths of both lists by iterating through them once. If one list is longer, advance its head pointer by the difference in lengths so both pointers have the same remaining distance to the tail. After alignment, move both pointers forward simultaneously until they either meet at the intersection node or reach null. The core insight: once both pointers have equal distance left, any shared node must appear at the same step. This approach relies purely on pointer traversal within a linked list and uses constant memory.
Approach 2: Two Pointer Technique (O(n + m) time, O(1) space)
The optimal trick uses two pointers starting at the heads of each list. Traverse both lists normally. When a pointer reaches the end, redirect it to the head of the other list. After switching lists once, both pointers will traverse exactly n + m nodes. If an intersection exists, they meet at that node; otherwise both reach null simultaneously. The key insight is path equalization: switching heads cancels the length difference automatically. This elegant method is a classic application of the two pointers pattern on a linked list structure.
Some engineers initially attempt a solution using a hash table, storing visited nodes from one list and checking membership while scanning the other. That works but requires O(n) extra space, which is unnecessary given the pointer-based solutions above.
Recommended for interviews: The two pointer switching technique is the solution most interviewers expect. It demonstrates strong pointer reasoning and avoids extra memory. Explaining the length-difference approach first can show structured thinking, but the two pointer method signals deeper familiarity with linked list traversal patterns.
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.
This solution first determines the lengths of both lists. It advances the pointer for the longer list by the difference in lengths. Then, it compares nodes in both lists step by step to find the intersection.
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Time Complexity: O(m + n).
Space Complexity: O(1).
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.
This implementation uses two pointers initialized to the heads of the two lists. Each pointer traverses its list and switches to the other list once it reaches the end. They meet at the intersection node or at null if there is none.
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Java
Python
C#
JavaScript
Time Complexity: O(m + n).
Space Complexity: O(1).
| Approach | Complexity |
|---|---|
| Approach 1: Using Length Difference | Time Complexity: O(m + n). |
| Approach 2: Two Pointer Technique | Time Complexity: O(m + n). |
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Length Difference Alignment | O(n + m) | O(1) | When you want a straightforward pointer alignment method that explicitly handles list length differences. |
| Two Pointer Switching | O(n + m) | O(1) | Best general solution. Automatically equalizes path lengths without computing sizes. |
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