[算法]回文检测

1.找链表的中点位置，并把中点以前的各个结点的值存入到栈中。
2.针对偶数或者奇数个链表结点，对中点结点做一个小的调整。
3.依次对比栈中结点的值后链表后半部分结点的值是否相等，决定是否是回文结构。

bool isPalindrome(ListNode* head) 
{
	bool isPalindrome = true;
	if(head == NULL)
	{
		return true;
	}
	
	stack<int> s;
	
	//找到链表的中点
	ListNode *fast = head;
	ListNode *low = head;
	while(fast != NULL && fast->next != NULL)
	{
		s.push(low->val);
		fast = fast->next->next;
		low = low->next;
	}//while
	if(fast != NULL)
	{
		low = low->next;
	}
	
	//对比前后两段链表的内容
	while(low != NULL)
	{
		if(low->val != s.top())
		{
			isPalindrome = false;
			break;
		}
		low = low->next;
		s.pop();
	}//while
	
	return isPalindrome;
}

bool isPalindrome(ListNode* head) 
{
	if(head == NULL || head->next == NULL)
		return true;
		
	ListNode *fast = head;
	ListNode *low = head;
	
	//利用快慢指针找到链表的中点
	while(fast != NULL && fast->next != NULL)
	{
		fast = fast->next->next;
		low = low->next;
	}
	if(fast != NULL)
	{
		low = low->next;
	}
	
	//逆序链表的后半部分
	ListNode dummy(-1);
	dummy.next = low;
	while(low->next != NULL)
	{
		ListNode *tmp = low->next;
		low->next = tmp->next;
		tmp->next = dummy.next;
		dummy.next = tmp;
	}//while
	
	//对比前后两部分
	low = dummy.next;
	fast = head;
	while(low != NULL)
	{
		if(fast->val != low->val)
			return false;
		low = low->next;
		fast = fast->next;
	}
	
	return true;
}