Write a program to find the node at which the intersection of two singly linked lists begins.

For example, the following two linked lists:

A: a1 → a2 ↘ c1 → c2 → c3 ↗ B: b1 → b2 → b3

begin to intersect at node c1.

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.

**Understand the problem:**

The key to understand the problem is when two linked lists have intersections, they will start "merging".

The solution is first to calculate the length of the list. Move the head of the longer list by (lenA - lenB) steps. Then compare each node of the two lists. If equal, the intersection point was found. If reached to the end, means null.

**Code (Java):**

/** * Definition for singly-linked 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) { if (headA == null || headB == null) { return null; } int lenA = getListLength(headA); int lenB = getListLength(headB); ListNode pA = headA; ListNode pB = headB; if (lenA > lenB) { for (int i = 0; i < lenA - lenB; i++) { pA = pA.next; } } else if (lenA < lenB) { for (int i = 0; i < lenB - lenA; i++) { pB = pB.next; } } while (pA != null && pB != null) { if (pA == pB) { return pA; } pA = pA.next; pB = pB.next; } return null; } private int getListLength(ListNode head) { int len = 0; ListNode p = head; while (p != null) { len++; p = p.next; } return len; } }

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