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

方法1：现在想到的，最直接也是最好理解的一种方法 

/**
 * 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) {
        List<Integer> list=new ArrayList<>();
        while(headA != null){
            list.add(headA.val);
            headA=headA.next;
        }
        while(headB!=null){
            if(list.contains(headB.val)){
                return headB;
            }
            headB=headB.next;
        }
        return null;       
    }
}

时间复杂度：O(n)

空间复杂度：O(n)