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 * 104
1 <= Node.val <= 105
0 <= skipA <= m
0 <= skipB <= n
intersectVal
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?Given the heads of two singly linked lists, headA
and headB
, find the node at which the two lists intersect. If there is no intersection, return null
. The linked lists retain their original structure after the function returns.
This problem can be efficiently solved using the two-pointer technique. The core idea is to traverse both lists simultaneously, switching to the other list when one pointer reaches the end.
Algorithm:
ptrA
and ptrB
, to the heads of headA
and headB
, respectively.ptrA
and ptrB
point to the same node (intersection) or both become null
(no intersection).ptrA
is null
, set it to headB
.ptrA
to the next node (ptrA.next
).ptrB
is null
, set it to headA
.ptrB
to the next node (ptrB.next
).ptrA
(or ptrB
, they will be the same if an intersection exists).Time Complexity: O(m + n), where m and n are the lengths of the two linked lists. Each list is traversed at most once.
Space Complexity: O(1), as we only use a constant amount of extra space for the pointers.
Here's the code implementation in several languages:
Python:
class ListNode:
def __init__(self, x):
self.val = x
self.next = None
class Solution:
def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode:
ptrA = headA
ptrB = headB
while ptrA != ptrB:
ptrA = ptrA.next if ptrA else headB
ptrB = ptrB.next if ptrB else headA
return ptrA
Java:
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution {
public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
ListNode ptrA = headA;
ListNode ptrB = headB;
while (ptrA != ptrB) {
ptrA = ptrA == null ? headB : ptrA.next;
ptrB = ptrB == null ? headA : ptrB.next;
}
return ptrA;
}
}
C++:
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
ListNode *ptrA = headA;
ListNode *ptrB = headB;
while (ptrA != ptrB) {
ptrA = ptrA ? ptrA->next : headB;
ptrB = ptrB ? ptrB->next : headA;
}
return ptrA;
}
};
JavaScript:
/**
* 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) {
let ptrA = headA;
let ptrB = headB;
while (ptrA !== ptrB) {
ptrA = ptrA === null ? headB : ptrA.next;
ptrB = ptrB === null ? headA : ptrB.next;
}
return ptrA;
};
These code examples all implement the same core algorithm, demonstrating its elegance and efficiency across different programming languages. The conditional assignment (ptrA = ptrA ? ptrA->next : headB;
in C++, for example) handles the case where a pointer reaches the end of its list.