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.

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.

可以将A,B两个链表看做两部分，交叉前与交叉后。例子中

交叉前A：          a1 → a2

交叉前B:   b1 → b2 → b3

交叉后AB一样： c1 → c2 → c3

所以 ，交叉后的长度是一样，所以，交叉前的长度差即为总长度差。

只要去除长度差，距离交叉点就等距了。

 public class Solution {

     public ListNode getIntersectionNode(ListNode headA, ListNode headB) {

         int lenA=0 ,lenB=0;

         int dis;

         for(ListNode x = headA;x != null;x = x.next) lenA++; //len of a

         for(ListNode x = headB;x != null;x = x.next) lenB++;// len of b

         dis = lenB - lenA;//distence of a and b

         if(dis > 0)//b is longer than a  move headb

             for(int i = dis;i > 0;i --) headB = headB.next;

         else  // a is longer than b move heada

             for(int i = -dis;i > 0;i --) headA = headA.next;

         while(headA != headB){ // heada and headb are equal?

             headA = headA.next;

             headB = headB.next;

         }

         return headA;

     }

参考：

http://www.cnblogs.com/ganganloveu/p/4128905.html