public boolean chkPalindrome(ListNode head) {
        if(head == null) return false;//判斷一下鏈表是不是空，空的話直接返回false
        ListNode fast = head;//快指針fast，初始等于head
        ListNode slow = head;//慢指針slow，初始等于head
        while(fast != null && fast.next != null){//如果鏈表是奇數，fast.next == null停下，如果鏈表是偶數fast == null停下
            fast = fast.next.next;//fast走兩步
            slow = slow.next;//slow走一步
 
        }
        ListNode cur = slow.next;//cur等于slow的下一個節(jié)點
        while(cur != null){//cur不為空開始反轉
            ListNode curNext = cur.next;//curNext等于cur的下一個節(jié)點
            cur.next = slow;//開始反轉
            slow = cur;//反轉完了后讓slow等于cur
            cur = curNext;//cur再往后走一步。
        }
        while(head != slow){//判斷是不是回文結構
            if(head.val != slow.val){//不是回文結構
                return false;
            }
            if(head.next == slow){//偶數鏈表的情況
                return true;
            }
            head = head.next;
            slow = slow.next;
        }
        return true;
    }

 public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        if(headB == null || headA == null) return null;//判斷其中一條鏈表的頭節(jié)點為空就是沒有焦點
        ListNode ps = headA;
        ListNode pl = headB;
        int lenA = 0;
        int lenB = 0;
        while(ps != null){求出ps鏈表的長度
            lenA++;
            ps = ps.next;
        }
        ps = headA;//讓ps重新等于頭節(jié)點
        while(pl != null){求出pl鏈表的長度
            lenB++;
            pl = pl.next;
        }
        pl = headB;//讓pl重新等于頭節(jié)點
        int len = lenA - lenB;
        if(len < 0){//判斷ps長還是pl長
            ps = headB;
            pl = headA;
            len = lenB - lenA;
        }
         while(len != 0)//求兩條鏈表的差值
                ps = ps.next;
                len--;
 
            }
        while(ps != pl){
            
            ps = ps.next;
            pl = pl.next;
        }
        return ps;
    }