题解 | #判断一个链表是否为回文结构#

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 *	ListNode(int x) : val(x), next(nullptr) {}
 * };
 */
#include <stack>
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param head ListNode类 the head
     * @return bool布尔型
     */
    bool isPail(ListNode* head) {
        // write code here
        if (head == nullptr || head->next == nullptr) {
            return true;
        }
        // 快慢指针找中点
        ListNode* slow = head;
        ListNode* fast = head->next;
        while (fast->next != nullptr && fast->next->next != nullptr) {
            slow = slow->next;
            fast = fast->next->next;
        }
        
        ListNode* headRight = slow->next;

        // 左边列表压栈: 奇数(fast->next != nullptr)时headRight也压栈，偶数(fast->next != nullptr)时则否
        slow = head;
        stack<int> myStack;
        while (slow != headRight) {
            myStack.push(slow->val);
            slow = slow->next;
        }
        if (fast->next != nullptr) {  
            myStack.push(headRight->val);
        }
  
        // 取栈顶与右边比较
        while (headRight != nullptr) {
            if(headRight->val != myStack.top()) {
                return false;
            } else {
                myStack.pop();
                headRight = headRight->next;
            }
        }
        return true;
    }
};