LeetCode OJ 160. 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.

这个题目是要找出两个链表的交叉点，该如何解决呢？如果两个链表相交，我们从A链表出发移动一段距离alen，出B列表出发移动一段距离blen，那么会发现他们指向同一个节点c1。那么这个距离是多少呢？

我们把每一个链表看成两段，不相交的一段和相交的一段。相交的一段对于两个链表长度是一样的，不想交的一段链表的长度是不同的。如果我们分别计算出两个链表的长度，然后计算他们长度的差值f，然后在较长的链表上先移动距离f，再同时从A,B链表出发开始遍历，若发现他们指向同一个节点，则他们相交并返回相交的点，若他们不想交，则会遍历到链表的尾部，则返回null。代码如下：

 /**
  * 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) {
         ListNode p1 = headA, p2 = headB;
         int len1 = 0, len2 = 0;
         while (p1 != null) {    //求链表A的长度
             p1 = p1.next;
             len1++;
         }
         while (p2 != null) {    //求链表B的长度
             p2 = p2.next;
             len2++;
         }
         p1 = headA;
         p2 = headB;
         if (len1 > len2) {      //计算链表长度的差值并在较长的链表上向后移动|len1-len2|
             for (int i = 0;i < len1 - len2; i++) {
                 p1 = p1.next;
             }
         } else {
             for (int i = 0;i < len2 - len1; i++) {
                 p2 = p2.next;
             }
         }
         while (p1 != p2) {      //向后遍历链表A和链表B，找到相交的节点，若遍历到最后，返回null
             p1 = p1.next;
             p2 = p2.next;
         }
         return p1;
     }
 }