程序员成长之旅——链表的回文结构-优快云博客

本文链接：https://blog.youkuaiyun.com/wuweiwuju___/article/details/97009289

本文介绍了三种使用C++实现链表回文结构检测的方法：通过快慢指针找到中点并反转后半部分进行对比；将链表元素存入数组进行回文验证；计算链表长度后反转后半部分再逐个元素比较。

程序员成长之旅——链表的回文结构

- 题目简介
- 牛客网

题目简介

在这里插入图片描述

牛客网

/*
struct ListNode {
    int val;
    struct ListNode *next;
    ListNode(int x) : val(x), next(NULL) {}
};*/
class PalindromeList {
public:
    bool chkPalindrome(ListNode* A) {
        // write code here
        if(A == NULL && A->next == NULL)
            return true;
        ListNode* fast = A;
        ListNode* slow = A;
        //找中间结点
        while(fast)
        {
            slow = slow->next;
            fast = slow->next;
        }
        ListNode* cur = slow->next;
        ListNode* head = slow;
        //反转后面节点
        while(cur)
        {
            ListNode* prev = cur->next;
            cur->next = head;
            head->next = NULL;
            head = cur;
            cur = prev;
        }
        //比较
        while(head)
        {
            if(head->val != A->val)
            {
               return false;
            }     
               head = head->next;
               A = A->next;
            
        }
        return true;
    }
};

/*
struct ListNode {
    int val;
    struct ListNode *next;
    ListNode(int x) : val(x), next(NULL) {}
};*/
class PalindromeList {
public:
    bool chkPalindrome(ListNode* A) {
        // write code here
        int a[900];
        int n = 0;
        struct ListNode* cur = A;
        while(cur)
        {
            a[n++] = cur->val;
            cur = cur->next;
        }
        
        int left = 0, right = n-1;	
        while(left < right)
        {
              if(a[left] != a[right])
                  return false;
            ++left;
            --right;
        }
        
        return true;
    }
};

/*
struct ListNode {
    int val;
    struct ListNode *next;
    ListNode(int x) : val(x), next(NULL) {}
};*/
class PalindromeList {
public:
    struct ListNode* reverseList(struct ListNode* head) {
        struct ListNode* newhead = NULL;
        struct ListNode* cur = head;
        while(cur)
        {
            struct ListNode* next = cur->next;
            cur->next = newhead;
            newhead = cur;
            cur = next;
        }
     
        return newhead;
    }
     
    bool chkPalindrome(ListNode* A) {
        int n = 0;
        struct ListNode* cur = A;
        while(cur)
        {
            cur = cur->next;
            ++n;
        }
         
        cur = A;
        int mid = n/2;
        while(mid--)
        {
            cur = cur->next;
        }
         
        struct ListNode* head1 = A;
        struct ListNode* head2 = reverseList(cur);
         
        mid = n/2;
        while(mid--)
        {
            if(head1->val == head2->val)
            {
                head1 = head1->next;
                head2 = head2->next;
            }
            else
            {
                   return false;
            }
        }
         
        return true;
    }
};