难度：Easy
解题思路：先找到链表中间节点，然后切段，然后把后一段翻转，最后对比。
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:
bool isPalindrome(ListNode* head) {
ListNode *dummy = new ListNode(-1);
dummy->next = head;
ListNode *slow = dummy;
ListNode *fast = dummy;
while(fast->next != NULL && fast->next->next != NULL)
{
slow = slow->next;
fast = fast->next->next;
}
ListNode *new_head = NULL;
if(fast->next != NULL)
{
new_head = slow->next->next;
slow->next = NULL;
}
else
{
new_head = slow->next;
slow->next = NULL;
}
new_head = reverseList(new_head);
ListNode *first_cur = dummy->next;
ListNode *second_cur = new_head;
while(first_cur != NULL && second_cur != NULL)
{
if(first_cur->val != second_cur->val)
return false;
else
{
first_cur=first_cur->next;
second_cur = second_cur->next;
}
}
return true;
}
ListNode* reverseList(ListNode* head) {
ListNode* pre = NULL;
ListNode* cur = head;
while(cur != NULL)
{
ListNode *ne = cur->next;
cur->next = pre;
pre = cur;
cur = ne;
}
return pre;
}
};
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运行结果：23ms，超过56.02%
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