OR36 链表的回文结构

乌龟跌倒2024-01-26 20:35

一、思路

二、代码

一、思路

找到中间节点

后半部分逆置链表

定义两个指针，一个从头开始出发

一个从中间位置开始出发

但是注意：链表个数可能是奇数或者偶数，需要注意中间节点的计算

二、代码

cs 复制代码

struct ListNode* reverseList(struct ListNode* head) {
    if(head == NULL)
    {
        return NULL;
    }
    struct ListNode *prev = NULL,*cur = head, *next = cur->next;
    while(next)
    {
        if(prev == NULL)
        {
            cur->next = NULL;
        }
        else
        {
            cur->next = prev;
        }
        prev = cur;
        cur = next;
        next = next->next;
    }
    cur->next = prev;
    return cur;
}
#include <asm-generic/errno.h>
class PalindromeList {
public:
    bool chkPalindrome(ListNode* A) {
        // write code here
        //找中间节点
        struct ListNode *slow = A, *fast = A;
        while(fast && fast->next)
        {
            slow = slow->next;
            fast = fast->next->next;
        }
        struct ListNode *mid = slow;
       struct ListNode *rhead = reverseList(slow);

        //比较
        struct ListNode *begin = A;
        while(begin && mid)
        {
            if(begin->val != rhead->val)
                return false;
                begin = begin->next;
                mid = mid->next;
        }
        return true;
    }
};