234. 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?
题目大意：
判断一个单链表是否为回文链表。
思路：
找到链表中间的节点，将链表从中间分为2部分，右半部分进行链表反向转换，然后左半部分和反转后的右半部分链表进行比较。得出结果。
代码如下：
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/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class
Solution {
public
:
ListNode * reverseList(ListNode *head)
//链表反转
{
ListNode *pre,*next;
pre = NULL;
next = NULL;
while
(head)
{
next = head->next;
head->next = pre;
pre = head;
head = next;
}
return
pre;
}
bool
isPalindrome(ListNode* head) {
if
( NULL == head || NULL == head->next)
return
true
;
int
len = 0;
ListNode *p = head;
while
(p)
{
len++;
p = p->next;
}
ListNode * rightListHead;
rightListHead = head;
int
leftLen = len / 2;
int
rightLen = len - leftLen;
int
i = leftLen;
while
(i)
{
rightListHead = rightListHead->next;
i--;
}
rightListHead = reverseList(rightListHead);
ListNode * left = head;
ListNode * right = rightListHead;
while
(i < leftLen)
{
if
(left->val == right->val)
{
left = left->next;
right = right->next;
}
else
{
return
false
;
}
i++;
}
return
true
;
}
};
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复习了单链表反转的方法。
本文转自313119992 51CTO博客，原文链接:http://blog.51cto.com/qiaopeng688/1837428
