Intersection of Two Linked Lists
Problem Description
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 is0if there is no intersected node.listA- The first linked list.listB- The second linked list.skipA- The number of nodes to skip ahead inlistA(starting from the head) to get to the intersected node.skipB- The number of nodes to skip ahead inlistB(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:
- The number of nodes of
listAis in them. - The number of nodes of
listBis in then. 1 <= m, n <= 3 * 1041 <= Node.val <= 1050 <= skipA <= m0 <= skipB <= nintersectValis0iflistAandlistBdo not intersect.intersectVal == listA[skipA] == listB[skipB]iflistAandlistBintersect.
Follow up: Could you write a solution that runs in O(m + n) time and use only O(1) memory?
Solve it on LeetCode
Approaches
1. Brute Force
Intuition:
The simplest method to determine if the linked lists intersect is to compare each node in one list with every node in the other list. If there is a node that matches, then that node is the intersection node.
Code:
- Time Complexity: O(m * n) where m and n are the lengths of the two linked lists.
- Space Complexity: O(1) as we are not using any extra space except two pointers.
2. HashSet
Intuition:
We can use a hash set to store all the nodes of one of the linked lists, then traverse the second linked list to see if any node is already in the set. If a node is found in the set, that node is the intersection node.
Code:
- Time Complexity: O(m + n) where m and n are the lengths of the two linked lists.
- Space Complexity: O(n) where n is the length of the longer linked list since all its nodes are stored in a set.
3. Two Pointers
Intuition:
The most optimal approach uses two pointers. Initially, set two pointers to the heads of the two linked lists. Traverse through the linked lists, and when a pointer reaches the end of a linked list, redirect it to the head of the other linked list. If the lists intersect, the two pointers will eventually converge at the intersection node after (m + n) - c steps, where c is the length of the shared tail. If they don't intersect, both pointers will eventually become null, and the loop will end.
Code:
- Time Complexity: O(m + n) where m and n are the lengths of the two linked lists.
- Space Complexity: O(1) as no extra space is used, just pointers are moved.