public ListNode reverseStore(ListNode head) {
        // 初始化，讓快指針和慢指針都處于鏈表的頭部
        ListNode slow=head;
        ListNode fast=head;
        while(fast!=null&&fast.next!=null){//如果快指針并且下一個(gè)不為空
            fast=fast.next.next;//快指針移動(dòng)兩個(gè)
            slow=slow.next;//慢指針移動(dòng)一個(gè)       
        }
        return slow;
  }

public boolean hasCycle(ListNode head) {
     if (head == null)
        return false;
     //快慢兩個(gè)指針，初始時(shí)都指向鏈表的頭結(jié)點(diǎn)
     ListNode slow = head;
     ListNode fast = head;
     while (fast != null && fast.next != null) {
         //慢指針每次走一步
         slow = slow.next;
        //快指針每次走兩步
        fast = fast.next.next;
        //如果相遇，說明有環(huán)，直接返回true
        if (slow == fast)
           return true;
    }
    //否則就是沒環(huán)
    return false;
}

//刪除鏈表倒數(shù)第n個(gè)節(jié)點(diǎn)
public ListNode removeBackEndNode(ListNode head, int n) {
        if (head == null || head.next == null) {
            return null;
        }
​
        ListNode dummyHead = new ListNode(-1);
        dummyHead.next = head;
       //初始時(shí)讓快慢指針都指向鏈表的頭部
        ListNode slow = dummyHead;
        ListNode fast = dummyHead;
​
        for (int i = 0; i < n + 1; i++) {
            fast = fast.next;
        }
​
        while (fast != null) {
            slow = slow.next;
            fast = fast.next;
        }
​
        slow.next = slow.next.next;  //為了解決斷鏈問題
​
        return dummyHead.next;
    }

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isPalindrome(ListNode head) {
        if(head == null || head.next == null){
            return true;
        }
​
        ListNode pre = null;//指向前一個(gè)節(jié)點(diǎn)的指針
        ListNode slow = head;//慢指針
        ListNode fast = head;//快指針
​
        while(fast!=null&&fast.next!=null){
            fast = fast.next.next;
            ListNode next = slow.next;
            slow.next = pre;//修改慢指針走過的節(jié)點(diǎn)指向前一個(gè)節(jié)點(diǎn)
            pre = slow;
            slow = next;
        }
        if(fast != null){
            slow = slow.next;
            //當(dāng)長度為奇數(shù)是，慢指針需要再走一個(gè)單位
        }
        while(pre!=null) {
            //判斷是否為回文串
            if(pre.val!=slow.val){
                return false;
            }
            pre = pre.next;
            slow = slow.next;
        }
        return true;
    }
}