判断回文链表

Palindrome Linked List

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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?

/**********************************************************
* Author : oyjb
* Email : jbouyang@126.com
* Last modified : 2015-07-18 19:21
* Filename : PalindromeLinkedList.cpp
* Description : copy right for oyjb
* *******************************************************/
#include <iostream>
using namespace std;

/*
* Definition for singly-linked list
*/
struct ListNode
{
int val;
ListNode *next;
ListNode(int x) : val(x),next(NULL){}
};

class Solution
{
public:
/*--利用快慢指针的思想定位链表中间节点位置--*/
bool isPalindrome(ListNode *head)
{
if (head == NULL || head->next == NULL)
return true;

/*--slow为慢指针一次走一步，fast为快指针一次走两步--*/
ListNode *slow = head;
ListNode *fast = head;

/*--考虑到链表长度的奇偶性，最终的fast指针有两种停留位确定while循环结束的条件--*/
while (fast->next != NULL && fast->next->next != NULL)
{
slow = slow->next;
fast = fast->next->next;
}

/*--无论slow和fast停留在何处，slow指针的下一个位置开始均代表链表后半部分，因此统一处理--*/
ListNode *second = ReverseList(slow->next);

/*--此时，second指针代表后半部分链表的逆序，因此从head位置开始与second位置开始比较判断是否时回文--*/
ListNode *first = head;
while (second != NULL)
{
if (first->val != second->val)
return false;
first = first->next;
second = second->next;
}
return true;
}

/*--非递归的就地逆转单链表--*/
ListNode *ReverseList(ListNode *head)
{
if (head == NULL || head->next == NULL)
return head;
ListNode *current = head->next;
head->next = NULL; //断开head头节点，开始逆转
ListNode *nex;
while (current != NULL)
{
nex = current->next; //保存下一个节点
current->next = head; //逆转
head = current; //更新逆转后的头节点
current = nex;
}
return head;
}
};