（LinkedList）Intersection of Two Linked Lists

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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/**

 * 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;            //我用的是判断环的方法，碰撞点开始到连接点的长度=head到连接点的长度。
                                         //一种更好的方法是线计算出2个链表的长度再移动节点。

        ListNode moveA=headA;

        ListNode moveB=headB;

        ListNode lastA=null;

        while(moveA!=null && moveA.next!=null)  moveA=moveA.next;

        while(moveB!=null && moveB.next!=null)  moveB=moveB.next;

        if(moveA!=moveB) return null;

        else

        {

            lastA=moveA;

            moveA.next=headB;

        }

        ListNode slow=headA.next;

        ListNode fast=headA.next.next;

        while(slow!=fast){

            slow=slow.next;

            fast=fast.next.next;

        }

        moveB=slow;

        moveA=headA;

        while(moveA!=moveB){

            moveA=moveA.next;

            moveB=moveB.next;

        }

        lastA.next=null;

        return moveB;

    }

}

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.