检查单向链表是否为回文 C

Question

我正在尝试检查单向链表是否为回文。约束是 - 算法必须在线性时间和恒定空间内。

我使用的基本算法如下 -

1. 使用快慢指针将列表分成两半。
2. 原地逆转下半场。
3. 比较上半场和下半场。
4. 构造回原始列表
5. 返回结果。

我的实现适用于列表中元素数量为偶数但元素数量为奇数时失败。

/*
 * @brief   Checks if a list is a palindrome or not
 */
bool is_palindrome(node *head)
{
    node *first;                /* Points to head node of 1st half */
    node *second;               /* Points to head node of 2nd half */
    node *f_ptr = head;         /* Fast pointer */
    node *s_ptr = head;         /* Slow pointer */
    node *prev = NULL;          /* Previous to slow pointer */
    bool ret = false;           /* Return value */

    while (f_ptr && f_ptr->next && f_ptr->next->next)
    {
        prev = s_ptr;
        s_ptr = s_ptr->next;
        f_ptr = f_ptr->next->next;
    }

    /* List with even number of elements */
    if (!(f_ptr->next->next))
    {
        first = head;
        second = s_ptr->next;
        s_ptr->next = NULL;
        /* Reverse the second half */
        second = reverse_list(&second);
        /* Compare the first & second half */
        ret = are_identical(first, second);
        /* Finally, construct the original list back */
        second = reverse_list(&second);
        s_ptr->next = second;
        print_list(head);
    }
    /* List with odd number of elements */
    if (!(f_ptr->next))
    {
        first = head;
        second = s_ptr->next;
        prev->next = NULL;
        s_ptr->next = NULL;
        /* Reverse the second half */
        second = reverse_list(&second);
        /* Compare the first & second half */
        ret = are_identical(first, second);
        /* Finally, construct the original list back */
        second = reverse_list(&second);
        prev->next = s_ptr; s_ptr->next = second;
        print_list(head);
    }
    return ret;
}

有人可以帮我弄清楚我在处理奇数案件时做错了什么吗？可以找到此程序的完整实现 here .

Answer 1

感谢s7amuser指出错误。以下是适用于偶数和奇数情况的实现。

/*
 * @brief   Checks if a list is a palindrome or not
 */
bool is_palindrome(node *head)
{
    node *first;                /* Points to head node of 1st half */
    node *second;               /* Points to head node of 2nd half */
    node *f_ptr = head;         /* Fast pointer */
    node *s_ptr = head;         /* Slow pointer */
    node *prev = NULL;          /* Previous to slow pointer */
    bool ret = false;           /* Return value */

    while (f_ptr && f_ptr->next && f_ptr->next->next)
    {
        prev = s_ptr;
        s_ptr = s_ptr->next;
        f_ptr = f_ptr->next->next;
    }

    /* List with odd number of elements */
    if (!(f_ptr->next))
    {
        first = head;
        second = s_ptr->next;
        prev->next = NULL;
        s_ptr->next = NULL;
        /* Reverse the second half */
        second = reverse_list(&second);
        /* Compare the first & second half */
        ret = are_identical(first, second);
        /* Finally, construct the original list back */
        second = reverse_list(&second);
        prev->next = s_ptr; s_ptr->next = second;
        print_list(head);
    }
    /* List with even number of elements */
    else if (!(f_ptr->next->next))
    {
        first = head;
        second = s_ptr->next;
        s_ptr->next = NULL;
        /* Reverse the second half */
        second = reverse_list(&second);
        /* Compare the first & second half */
        ret = are_identical(first, second);
        /* Finally, construct the original list back */
        second = reverse_list(&second);
        s_ptr->next = second;
        print_list(head);
    }
    return ret;
}