Palindrome Linked List

Given a singly linked list, determine if it is a palindrome.

Follow up:

Could you do it in O(n) time and O(1) space?

判断是不是回文链表

<p style="box-sizing: border-box; margin-top: 0px; margin-bottom: 10px; color: rgb(51, 51, 51); font-family: 'Helvetica Neue', Helvetica, Arial, sans-serif; font-size: 14px; line-height: 30px;"></p>/**
 * 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) {
        vector<int> temp;
        while(head)
        {
            temp.push_back(head->val);
            head=head->next;
        }
        for(int i=0,j=temp.size()-1;i<j;i++,j--)
        {
            if(temp[i]!=temp[j])
            {
                return false;
            }
        }
    }
};<p style="box-sizing: border-box; margin-top: 0px; margin-bottom: 10px; color: rgb(51, 51, 51); font-family: 'Helvetica Neue', Helvetica, Arial, sans-serif; font-size: 14px; line-height: 30px;"></p>

/**
* 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) {
vector<int> temp;
while(head)
{
temp.push_back(head->val);
head=head->next;
}
for(int i=0,j=temp.size()-1;i<j;i++,j--)
{
if(temp[i]!=temp[j])
{
return false;
}
}
}
};

/**
* 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 == NULL || head->next == NULL) return true;
stack<int> temp;
ListNode* cur=head;
while(cur){
temp.push(cur->val);
cur=cur->next;
}
while(head){
if(head->val!=temp.top())
{
return false;
}
head=head->next;
temp.pop();
}
return true;

}
};

我们使用快慢指针找中点的原理是fast和slow两个指针，每次快指针走两步，慢指针走一步，等快指针走完时，慢指针的位置就是中点。我们还需要用栈，每次慢指针走一步，都把值存入栈中，等到达中点时，链表的前半段都存入栈中了，由于栈的后进先出的性质，就可以和后半段链表按照回文对应的顺序比较了。

class Solution {
public:
bool isPalindrome(ListNode* head) {
if (!head || !head->next) return true;
ListNode *slow = head, *fast = head;
stack<int> s;
s.push(head->val);
while (fast->next && fast->next->next) {
slow = slow->next;
fast = fast->next->next;
s.push(slow->val);
}
if (!fast->next) s.pop();
while (slow->next) {
slow = slow->next;
int tmp = s.top(); s.pop();
if (tmp != slow->val) return false;
}
return true;
}
};