/*
public class ListNode {
    int val;
    ListNode next = null;

    ListNode(int val) {
        this.val = val;
    }
}*/
public class Solution {
    public ListNode FindFirstCommonNode(ListNode pHead1, ListNode pHead2) {
         // 只要有⼀个为空,就不存在共同节点
        if (pHead1 == null || pHead2 == null) {
            return null;
        }
        ListNode head1 = pHead1;
        ListNode head2 = pHead2;
        while (head1 !=head2) {
        // 如果下⼀个节点为空,则切换到另⼀个链表的头节点,否则下⼀个节点
            head1 = (head1 == null) ? pHead2 : head1.next;
            head2 = (head2 == null) ? pHead1 : head2.next;
        }
        return head1;
    }
}

解题思想:

* 方式一:set 是否包含

* 方式二:计算个数提前移动差值

* 方式三:拼接列表(推荐)