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Question
Formatted question description: https://leetcode.ca/all/160.html
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?
Algorithm
If the two chains are the same length, then the corresponding ones can be found one by one, so you only need to shorten the long list.
The specific algorithm is: traverse the two linked lists respectively to obtain the corresponding lengths. Then find the difference in length, move the longer linked list backward by the number of differences, and then compare them one by one.
Code

public class Intersection_of_Two_Linked_Lists { /** * Definition for singlylinked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { * val = x; * next = null; * } * } */ /** * 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) { int lenA = getLength(headA); int lenB = getLength(headB); // move headA and headB to the same start point while (lenA > lenB) { headA = headA.next; lenA; } while (lenA < lenB) { headB = headB.next; lenB; } // find the intersection until end while (headA != headB) { // same reference headA = headA.next; headB = headB.next; } return headA; } private int getLength(ListNode node) { int length = 0; while (node != null) { node = node.next; length++; } return length; } } } ############ /** * 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; } }

// OJ: https://leetcode.com/problems/intersectionoftwolinkedlists/ // Time: O(A + B) // Space: O(1) class Solution { int getLength(ListNode *h) { int len = 0; for (; h; h = h>next) ++len; return len; } public: ListNode *getIntersectionNode(ListNode *a, ListNode *b) { int la = getLength(a), lb = getLength(b); if (la < lb) swap(a, b), swap(la, lb); for (int i = la  lb; i > 0; i) a = a>next; while (a && b && a != b) a = a>next, b = b>next; 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 ############ class Solution: def getIntersectionNode(self, headA, headB): if headA is None or headB is None: return None pa = headA pb = headB while pa is not pb: pa = headB if pa is None else pa.next pb = headA if pb is None else pb.next return pa

/** * 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 } }