题解 | 链表的回文结构

/*
struct ListNode {
    int val;
    struct ListNode *next;
    ListNode(int x) : val(x), next(NULL) {}
};*/
//找中间节点，反转后续链表，比较
#include <csetjmp>
class PalindromeList {
  public:
    struct ListNode* reverseList(struct ListNode* head) {
        struct ListNode* newnode = NULL, *ptr = head, *nextnode, *ptr1;
        while (ptr) {
            nextnode = ptr->next;
            if (newnode == NULL) {
                newnode = ptr;
                newnode->next = NULL;
            } else {
                ptr1 = newnode;
                newnode = ptr;
                newnode->next = ptr1;
            }
            ptr = nextnode;
        }
        return newnode;
    }
    struct ListNode* middleNode(struct ListNode* head) {
        struct ListNode* ptr = head;
        struct ListNode* ptr1 = head;
        while (ptr != NULL) {
            if (ptr->next == NULL) {
                break;
            }
            ptr1 = ptr1->next;
            ptr = ptr->next->next;
        }
        return ptr1;
    }
    bool chkPalindrome(ListNode* A) {
        struct ListNode* ptr = middleNode(A);
        struct ListNode* ptr1 = reverseList(ptr);
        while (A && ptr1)
        {
            if (A->val != ptr1->val) {
                return false;
            }
            A = A->next;
            ptr1 = ptr1->next;
        }
        return true;
    }
};