234 Palindrome Linked List 回文链表

请检查一个链表是否为回文链表。

进阶：

你能在 O(n) 的时间和 O(1) 的额外空间中做到吗？

详见：https://leetcode.com/problems/palindrome-linked-list/description/

Java实现：

方法一：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isPalindrome(ListNode head) {
        //0个节点或是1个节点
        if(head==null||head!=null&&head.next==null){
            return true;
        }
        ListNode slow=head;
        ListNode fast=head;
        Stack<Integer> stk=new Stack<Integer>();
        stk.push(head.val);
        while(fast.next!=null&&fast.next.next!=null){
            slow=slow.next;
            fast=fast.next.next;
            stk.push(slow.val);
        }
        //链表长度为偶数
        if(fast.next!=null){
            slow=slow.next;
        }
        while(slow!=null){
            int tmp=stk.pop();
            if(slow.val!=tmp){
                return false;
            }
            slow=slow.next;
        }
        return true;
    }
}

方法二：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isPalindrome(ListNode head) {
        //0个节点或是1个节点
        if(head==null||head!=null&&head.next==null){
            return true;
        }
        ListNode slow=head;
        ListNode fast=head;
        ListNode first=null;
        while(fast!=null&&fast.next!=null){
            first=slow;
            slow=slow.next;
            fast=fast.next.next;
        }
        fast=first.next;
        first.next=null;
        ListNode pre=null;
        ListNode next=null;
        while(fast!=null){
            next=fast.next;
            fast.next=pre;
            pre=fast;
            fast=next;
        }
        first=head;
        fast=pre;
        while(first!=null&&fast!=null){
            if(first.val!=fast.val){
                return false;
            }
            first=first.next;
            fast=fast.next;
        }
        return true;
    }
}

C++实现：

方法一：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    bool isPalindrome(ListNode* head) {
        if(head==nullptr||head->next==nullptr)
        {
            return true;
        }
        ListNode *slow=head;
        ListNode *fast=head;
        stack<int> stk;
        stk.push(head->val);
        while(fast->next&&fast->next->next)
        {
            slow=slow->next;
            fast=fast->next->next;
            stk.push(slow->val);
        }
        if(!fast->next)
        {
            stk.pop();
        }
        while(slow->next)
        {
            slow=slow->next;
            int tmp=stk.top();
            if(slow->val!=tmp)
            {
                return false;
            }
            stk.pop();
        }
        return true;
    }
};

方法二：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    bool isPalindrome(ListNode* head) {
        if(!head||!head->next)
        {
            return true;
        }
        ListNode *slow=head;
        ListNode *fast=head;
        ListNode *first=nullptr;
        while(fast&&fast->next)
        {
            first=slow;
            slow=slow->next;
            fast=fast->next->next;
        }
        fast=first->next;
        first->next=nullptr;
        ListNode *pre=nullptr;
        ListNode *next=nullptr;
        while(fast)
        {
            next=fast->next;
            fast->next=pre;
            pre=fast;
            fast=next;
        }
        first=head,fast=pre;
        while(first&&fast)
        {
            if(first->val!=fast->val)
            {
                return false;
            }
            first=first->next;
            fast=fast->next;
        }
        return true;
    }
};