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.

思路：先计算出两个链表的长度，然后算出两个链表的长度差diff，长的那个链表从前往后走diff步，然后开始比较两个链表对应结点的值，如果相等就是所要找的结点。

/** * 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) {        int lengthA=0;        int lengthB=0;        if(headA==null||headB==null) return null;         ListNode  pa=headA;        ListNode pb=headB;                while(pa.next!=null)              { lengthA++;                pa=pa.next;              }              lengthA++;            while(pb.next!=null)              { lengthB++;                pb=pb.next;              }              lengthB++;               if(pa!=pb) return null;            int diff=lengthA-lengthB;            if(diff>0)            {                while(diff>0)                {                    headA=headA.next;                    diff--;                }            }else{                while(diff<0)                {                    headB=headB.next;                    diff++;                }            }            while(headA!=null && headB!=null)            {                if(headA.val==headB.val) break;                else                   {                       headA=headA.next;                      headB=headB.next;                  }            }         return headA;     }}