LeetCode 160. Intersection of Two Linked Lists
Question
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
.
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 is0
if 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:
1 | Input: intersectVal = 8, listA = [4,1,8,4,5], listB = [5,6,1,8,4,5], skipA = 2, skipB = 3 |
Example 2:
1 | Input: intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2 |
Constraints:
- The number of nodes of
listA
is in them
. - The number of nodes of
listB
is in then
. 1 <= m, n <= 3 * 104
1 <= Node.val <= 105
0 <= skipA < m
0 <= skipB < n
intersectVal
is0
iflistA
andlistB
do not intersect.intersectVal == listA[skipA] == listB[skipB]
iflistA
andlistB
intersect.
Follow up: Could you write a solution that runs in O(m + n)
time and use only O(1)
memory?
Source: https://leetcode.com/problems/intersection-of-two-linked-lists/
Solution
1 | public ListNode getIntersectionNode(ListNode headA, ListNode headB) { |
LeetCode 160. Intersection of Two Linked Lists
http://yenotes.org/2022/03/09/LeetCode-160-Intersection-of-Two-Linked-Lists/