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 <bits/types/struct_tm.h>
#include <cstddef>
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
     */
    ListNode* reverse(ListNode* head) {
        ListNode* newhead = new ListNode(0);
        ListNode* begin = NULL;
        while (head) {
            ListNode* tmp = head->next;
            newhead->next = head;
            head->next = begin;
            begin = head;
            head = tmp;
        }
        return newhead->next;
    }
    bool isPail(ListNode* head) {
        if (!head) {
            return false;
        }
        ListNode* left = head;
        ListNode* right = head;
        ListNode *pre = NULL;
        int len = 0;
        while (head) {
            len++;
            head = head->next;
        }
        if (len==1) {
            return true;
        }
        int mid = len / 2;
        for (int i = 0; i < mid; i++) {
            pre = right;
            right = right->next;
        }
        pre->next = NULL;
        if (len % 2) {
            right = right->next;
            right = reverse(right);
            while (left) {
                if (left->val != right->val) return false;
                left = left->next;
                right = right->next;
            }
        }
        if (!(len % 2)) {
            right = reverse(right);
            while (left) {
                if (left->val != right->val) return false;
                left = left->next;
                right = right->next;
            }
        }
        return true;
        // write code here
    }
};

第一次做

看了思路还是挺简单的