@TOC

1. 题目

题目链接：链表的回文结构

2. 思路及小注意

这道题目整合了之前几道题目，返回链表中间节点，反转单链表。需要比较熟悉。
判断是否为回文链表 ——
:black_heart: 1. 找到中间节点
:black_heart: 2. 逆置后半个链表
:black_heart: 3. 遍历比较，确定终止条件

3. 题解

class PalindromeList {
public:
    bool chkPalindrome(ListNode* A) {
        // write code here
        // 1.寻找链表的中间节点
        struct ListNode* fast = A;
        struct ListNode* slow = A;
        while(fast != NULL && fast->next != NULL)
        {
            slow = slow->next;
            fast = fast->next->next;
        }
        // 2.反转链表
        struct ListNode* prev = NULL;
        struct ListNode* cur = slow;
        struct ListNode* next = slow->next;
        while(cur)
        {
            cur->next = prev;//反转
            //迭代
            prev = cur;
            cur = next;
            if(next)
                next = next->next;
        }
        // 3.比较是否相同
        while(A && prev)
        {
            if(A->val != prev->val)
            {
                return false;
            }
            A = A->next;
            prev = prev->next;
        }
        return true;
    }
};

:star: After all this time?
:black_heart: Always.

持续更新~@边通书