方法一：两个指针以同样速度走完相同长度路径必会相遇

public ListNode FindFirstCommonNode(ListNode pHead1, ListNode pHead2) {
        //if(pHead1==null||pHead2==null)return null;
        ListNode cur1 = pHead1;
        ListNode cur2 = pHead2;
        while(cur1!=cur2){
            //相遇时循环停止
            if(cur1==null){
                //1走完就去走2
                cur1 = pHead2;
            }else{
                //只有非空时才能往下一个位置移动！
                cur1 = cur1.next; 
            }
            if(cur2==null){
                //2走完转而走1
                cur2 = pHead1;
            }else{
                cur2 = cur2.next;   
            }
        }
        return cur1;
    }

方法二：双层循环，遍历搜索

嵌套循环，时间复杂度

public ListNode FindFirstCommonNode(ListNode pHead1, ListNode pHead2) {
        //两链表同时搜索，两个活动指针
        ListNode cur1 = pHead1;
        ListNode cur2 = pHead2;
        while(cur1!=null){
            //这一步非常重要，保证每次循环开始，指向头结点
            cur2 = pHead2;
            while(cur2!=null){
                if(cur1==cur2)return cur2;
                cur2 = cur2.next;
            }
            cur1 = cur1.next;
        }
        //若没有匹配结果，返回{}
        return null;
    }