LeetCode234. 回文链表(java)

题目：

请判断一个链表是否为回文链表。

示例：

代码：

• 解法一

/**
 * 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 mid=middleNode(head);
        ListNode root=reversal(mid);
        ListNode p1=head;
        ListNode p2=root;
        while(p2!=null&&p1.val==p2.val){
                p1=p1.next;
                p2=p2.next;
        }
        return p2==null?true:false;
    }
    private ListNode middleNode(ListNode head){	//寻找中心元素
        ListNode slow=head;
        ListNode fast=head.next;
        while(fast!=null&&fast.next!=null){
            slow=slow.next;
            fast=fast.next.next;
        }
        return slow.next;
    }
    //迭代实现
    private ListNode reversal(ListNode head){
        ListNode root=new ListNode(0);
        ListNode cur=root;
        while(head!=null){
            cur=head;
            head=head.next;
            cur.next=null;
            cur.next=root.next;
            root.next=cur;
        }
        return root.next;
    }
}

• 解法二

/**
 * 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 mid=middleNode(head);
        ListNode root=reversal(mid);
        ListNode p1=head;
        ListNode p2=root;
        while(p2!=null&&p1.val==p2.val){
                p1=p1.next;
                p2=p2.next;
        }
        return p2==null?true:false;
    }
    private ListNode middleNode(ListNode head){
        ListNode slow=head;
        ListNode fast=head.next;
        while(fast!=null&&fast.next!=null){
            slow=slow.next;
            fast=fast.next.next;
        }
        return slow.next;
    }
    //迭代实现
    private ListNode reversal(ListNode head){
        if(head==null||head.next==null){
            return head;
        }
        ListNode h=reversal(head.next);
        head.next.next=head;
        head.next=null;
        return h;
    }
}

• 别人的代码

 /**
 * 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;
        }
        if(head.next.next==null){
            return head.val == head.next.val;
        }
        ListNode slow = head;
        ListNode fast = head.next;
        
        while(fast.next!=null){
            if(slow.val == fast.next.val){
                if(fast.next.next!=null){
                    return false;
                }
                fast.next = null;
                slow = slow.next;
                fast = slow.next;
                if(fast==null || slow.val == fast.val){
                    return true;
                }
            }else{
                fast = fast.next;
            }
        }
        return false;
}
}