public boolean chkPalindrome(ListNode head) {
        if(head == null) return false;//判斷一下鏈表是不是空，空的話(huà)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(shù)，fast.next == null停下，如果鏈表是偶數(shù)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(gè)節(jié)點(diǎn)
        while(cur != null){//cur不為空開(kāi)始反轉(zhuǎn)
            ListNode curNext = cur.next;//curNext等于cur的下一個(gè)節(jié)點(diǎn)
            cur.next = slow;//開(kāi)始反轉(zhuǎn)
            slow = cur;//反轉(zhuǎn)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(jié)構(gòu)
            if(head.val != slow.val){//不是回文結(jié)構(gòu)
                return false;
            }
            if(head.next == slow){//偶數(shù)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(diǎn)為空就是沒(méi)有焦點(diǎn)
        ListNode ps = headA;
        ListNode pl = headB;
        int lenA = 0;
        int lenB = 0;
        while(ps != null){求出ps鏈表的長(zhǎng)度
            lenA++;
            ps = ps.next;
        }
        ps = headA;//讓ps重新等于頭節(jié)點(diǎn)
        while(pl != null){求出pl鏈表的長(zhǎng)度
            lenB++;
            pl = pl.next;
        }
        pl = headB;//讓pl重新等于頭節(jié)點(diǎn)
        int len = lenA - lenB;
        if(len < 0){//判斷ps長(zhǎng)還是pl長(zhǎng)
            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;
    }