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First, traverse both linked lists to determine their lengths. Calculate the difference in lengths and advance the pointer of the longer list by the length difference. Then move both pointers in tandem to find the intersection node.
Time Complexity: O(m + n).
Space Complexity: O(1).
1public class Solution {
2 public class ListNode {
3 public int val;
4 public ListNode next;
5 public ListNode(int x) { val = x; next = null; }
6 }
7
8 private int GetLength(ListNode head) {
9 int length = 0;
10 while (head != null) {
11 length++;
12 head = head.next;
13 }
14 return length;
15 }
16
17 public ListNode GetIntersectionNode(ListNode headA, ListNode headB) {
18 int lenA = GetLength(headA);
19 int lenB = GetLength(headB);
20
21 if (lenA > lenB) {
22 for (int i = 0; i < lenA - lenB; i++) headA = headA.next;
23 } else {
24 for (int i = 0; i < lenB - lenA; i++) headB = headB.next;
25 }
26
27 while (headA != headB) {
28 headA = headA.next;
29 headB = headB.next;
30 }
31 return headA;
32 }
33}The C# solution makes use of an auxiliary function to calculate the lengths of the linked lists, adjusts one of the heads if needed, and searches for the intersection node.
Use two pointers, each starting at the head of one list. Traverse the list until a pointer reaches null, then start traversing the other list from the beginning. Repeat until the two pointers meet at the intersection node.
Time Complexity: O(m + n).
Space Complexity: O(1).
1function
The JavaScript implementation uses two pointers that travel through both lists. They alternate between lists at the end of each traversal until they meet at the intersection point or return null if no intersection exists.