【nowcoder】链表的回文结构

牛客题目链接链表的回文结构
在这里插入图片描述

/*
struct ListNode {int val;struct ListNode *next;ListNode(int x) : val(x), next(NULL) {}
};*/
#include <cstdlib>
// 建议大伙自己对照我的代码画下图，假设A链表是：1 2 3 2 1
class PalindromeList {
public:bool chkPalindrome(ListNode* A) { // 空间复杂度O(1)// 找到中间节点 （slow是3）ListNode* slow = A, * fast = A;while (fast && fast->next) {slow = slow->next;fast = fast->next->next;}// 反转后半部分 //（3 2 1变成1 2 3，但是A并没有变成1 2 1 2 3，A这时是1 2 3，自己画下图就知道了）ListNode* cur = NULL, * next = NULL;while (slow) {next = slow->next;slow->next = cur;cur = slow;slow = next;}// 前半部分和后半部分比较 （next = A = 1 2 3，cur = 1 2 3）next = A;while (cur) {if (next->val != cur->val) {return false;}next = next->next;cur = cur->next;}return true;}bool chkPalindrome_(ListNode* A) { // 空间复杂度O(n)// 1.创建B链表，将A链表节点依次向B链表头插ListNode* B = NULL;ListNode* curA = A;while (curA) {ListNode* newNode = (ListNode*)malloc(sizeof(ListNode));newNode->val = curA->val;newNode->next = NULL;if (B == NULL) {B = newNode;}else {newNode->next = B;B = newNode;}curA = curA->next;}// 2.比较curA = A;ListNode* curB = B;while (curA) {if (curA->val != curB->val) {return false;}curA = curA->next;ListNode* del = curB;curB = curB->next;free(del);del = NULL;}return true;}
};