OJ.判断链表是否为回文结构

思路：先找到链表的中间节点，然后将链表从中间节点之后的后半段反转；然后将前半段与后半段逐一进行比较。

//链表的建立
struct ListNode {
int val;
struct ListNode *next;
};

bool isPalindrome(struct ListNode* head)
{
//找到中间节点
struct ListNodetwo=head,one=head;
while(two&&two->next)
{
one=one->next;
two=two->next->next;
}
//将后半段进行反转
struct ListNode newhead = NULL;
struct ListNode Next = NULL;
struct ListNode* cur = one;
while (cur)
{
Next = cur->next;
cur->next = newhead;
newhead = cur;
cur = Next;
}
//逐一比较，是回文结构返回true；不是则返回false。
while (head&&newhead)
{
if (head->val != newhead->val)
return false;
newhead = newhead->next;
head = head->next;
}
return true;
}