Problem Statement:
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 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.
Examples
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 <= nintersectValis 0 iflistAandlistBdo 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?
Approach
- Move both pointers one step at a time.
- When either pointer reaches the end, redirect it to the other list's head.
- If the lists intersect, the pointers will meet at the intersection node.
- If they don’t intersect, both will eventually become
nullafter traversing both lists.
We use two pointers pA and pB, starting at headA and headB.
Time Complexity:
Time Complexity = O(n + m)
Space Complexity:
Space Complexity = O(1)
Dry Run
Input: List A: 4 → 1 → 8 → 4 → 5, List B: 5 → 6 → 1 → 8 → 4 → 5
pA = 4, pB = 5 → not equal pA = 1, pB = 6 → not equal pA = 8, pB = 1 → not equal pA = 4, pB = 8 → not equal pA = 5, pB = 4 → not equal pA = null, pB = 5 → switch pA to headB pA = 5, pB = null → switch pB to headA pA = 6, pB = 4 → not equal pA = 1, pB = 1 → not equal pA = 8, pB = 8 → equal → exit loop
Output: Result: Node with value 8
Visualisation:
var getIntersectionNode = function(headA, headB) {
let pA = headA;
let pB = headB;
while(pA != pB) {
pA = pA == null ? headB : pA.next;
pB = pB == null ? headA : pB.next;
}
return pA;
};
def getIntersectionNode(headA, headB):
pA = headA
pB = headB
while pA != pB:
pA = headB if pA is None else pA.next
pB = headA if pB is None else pB.next
return pA
public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
ListNode pA = headA;
ListNode pB = headB;
while(pA != pB) {
pA = pA == null ? headB : pA.next;
pB = pB == null ? headA : pB.next;
}
return pA;
}
ListNode* getIntersectionNode(ListNode* headA, ListNode* headB) {
ListNode* pA = headA;
ListNode* pB = headB;
while(pA != pB) {
pA = pA == nullptr ? headB : pA->next;
pB = pB == nullptr ? headA : pB->next;
}
return pA;
}
struct ListNode* getIntersectionNode(struct ListNode* headA, struct ListNode* headB) {
struct ListNode* pA = headA;
struct ListNode* pB = headB;
while(pA != pB) {
pA = pA == NULL ? headB : pA->next;
pB = pB == NULL ? headA : pB->next;
}
return pA;
}
public ListNode GetIntersectionNode(ListNode headA, ListNode headB) {
ListNode pA = headA;
ListNode pB = headB;
while(pA != pB) {
pA = pA == null ? headB : pA.next;
pB = pB == null ? headA : pB.next;
}
return pA;
}
