Question
Formatted question description: https://leetcode.ca/all/160.html
160 Intersection of Two Linked Lists
Write a program to find the node at which the intersection of two singly linked lists begins.
For example, the following two linked lists:
begin to intersect at node c1.
Example 1:
Input: intersectVal = 8, listA = [4,1,8,4,5], listB = [5,0,1,8,4,5], skipA = 2, skipB = 3
Output: Reference of the node with value = 8
Input 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 [5,0,1,8,4,5]. There are 2 nodes before the intersected node in A; There are 3 nodes before the intersected node in B.
Example 2:
Input: intersectVal = 2, listA = [0,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1
Output: Reference of the node with value = 2
Input 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 [0,9,1,2,4]. 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: null
Input 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.
Notes:
If the two linked lists have no intersection at all, return null.
The linked lists must retain their original structure after the function returns.
You may assume there are no cycles anywhere in the entire linked structure.
Your code should preferably run in O(n) time and use only O(1) memory.
@taglinkedlist
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
Java

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; } } }

// 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; } };

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