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/
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;
}
};
参考：https://www.cnblogs.com/grandyang/p/4635425.html
234 Palindrome Linked List 回文链表
原文：https://www.cnblogs.com/xidian2014/p/8759212.html