Leetcode 234.Palindrome Linked List

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

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;
    }
};

O(n) O(1)

/**
 * 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;
        ListNode * fast=head;
        ListNode * slow = head;
        while(fast->next!=NULL && fast->next->next !=NULL){
            fast = fast->next->next;
            slow = slow->next;
        }
        ListNode* rev = reverse(slow->next);
        while(rev != NULL){
            if(rev->val != head->val) return false;
            rev=rev->next;
            head=head->next;
        }
        return true;

    }
    ListNode* reverse(ListNode * &head){
        ListNode * prev= NULL;
        while(head){
            ListNode * temp = head->next;
            head->next = prev;
            prev = head;
            head = temp;


        }
        return prev;
    }
};

REF:

• http://www.cnblogs.com/grandyang/p/4635425.html