（LinkedList）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:

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.