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160. Intersection of Two Linked Lists
Description
Given the heads of two singly linkedlists 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 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:
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 (2^{nd} node in A and 3^{rd} 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 (3^{rd} node in A and 4^{th} 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
listA
is in them
.  The number of nodes of
listB
is in then
. 1 <= m, n <= 3 * 10^{4}
1 <= Node.val <= 10^{5}
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?
Solutions

/** * Definition for singlylinked 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 a = headA, b = headB; while (a != b) { a = a == null ? headB : a.next; b = b == null ? headA : b.next; } return a; } }

/** * Definition for singlylinked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode* getIntersectionNode(ListNode* headA, ListNode* headB) { ListNode *a = headA, *b = headB; while (a != b) { a = a ? a>next : headB; b = b ? b>next : headA; } return a; } };

# Definition for singlylinked list. # class ListNode: # def __init__(self, x): # self.val = x # self.next = None class Solution: def getIntersectionNode(self, headA: ListNode, headB: ListNode) > ListNode: a, b = headA, headB while a != b: a = a.next if a else headB b = b.next if b else headA return a

/** * Definition for singlylinked list. * type ListNode struct { * Val int * Next *ListNode * } */ func getIntersectionNode(headA, headB *ListNode) *ListNode { a, b := headA, headB for a != b { if a == nil { a = headB } else { a = a.Next } if b == nil { b = headA } else { b = b.Next } } return a }

/** * Definition for singlylinked list. * class ListNode { * val: number * next: ListNode  null * constructor(val?: number, next?: ListNode  null) { * this.val = (val===undefined ? 0 : val) * this.next = (next===undefined ? null : next) * } * } */ function getIntersectionNode(headA: ListNode  null, headB: ListNode  null): ListNode  null { let a = headA; let b = headB; while (a != b) { a = a ? a.next : headB; b = b ? b.next : headA; } return a; }

/** * Definition for singlylinked list. * function ListNode(val) { * this.val = val; * this.next = null; * } */ /** * @param {ListNode} headA * @param {ListNode} headB * @return {ListNode} */ var getIntersectionNode = function (headA, headB) { let a = headA; let b = headB; while (a != b) { a = a ? a.next : headB; b = b ? b.next : headA; } return a; };

/** * Definition for singlylinked list. * public class ListNode { * public var val: Int * public var next: ListNode? * public init(_ val: Int) { * self.val = val * self.next = nil * } * } */ class Solution { func getIntersectionNode(_ headA: ListNode?, _ headB: ListNode?) > ListNode? { var a = headA var b = headB while a !== b { a = a == nil ? headB : a?.next b = b == nil ? headA : b?.next } return a } }