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.

Credits:

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Special thanks to @stellari for adding this problem and creating all test cases.

查找两个链表的共同节点

要求：时间复杂度为 O(N)，空间复杂度为 O(1)

设 A 的长度为 a + c， B 的长度为 b + c，其中 c 为尾部公共部分长度，可知 a + c + b

= b + c + a。

当访问 A 链表的指针访问到链表尾部时，令它从链表 B 的头部开始访问链表 B；同样

地，当访问 B 链表的指针访问到链表尾部时，令它从链表 A 的头部开始访问链表 A。这

样就能控制访问 A 和 B 两个链表的指针能同时访问到交点

/**
* 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) {
         ListNode A= headA;
        ListNode B= headB;
        while (A!=B) {
            if (A== null)  A = headB;
            else A= A.next;
            if (B==null)  B = headA;
            else B=B.next;
        }
       return A;
    
    }
}