C程序检查单链表是否为回文

给定一个单链字符列表，编写一个函数，如果给定列表是回文，则返回 true，否则返回 false。

方法1（通过反转列表）：
此方法需要 O(n) 时间和 O(1) 额外空间。
1）获取链表的中间。
2）反转链表的后半部分。
3)检查前半部分和后半部分是否相同。
4）通过再次反转后半部分并将其附加回前半部分来构造原始链表

要将列表分成两半，使用本文的方法 2。

当多个节点是偶数时，前半部分和后半部分正好包含半个节点。这种方法的挑战在于处理节点数为奇数的情况。我们不希望中间节点作为列表的一部分，因为我们将比较它们是否相等。对于奇怪的情况，我们使用单独的变量“中间节点”。

C

// C++ program to check if a linked list
// is palindrome
#include 
#include 
#include 
 
// Link list node
struct Node
{
    char data;
    struct Node* next;
};
 
void reverse(struct Node**);
bool compareLists(struct Node*,
                  struct Node*);
 
// Function to check if given linked
// list is palindrome or not
bool isPalindrome(struct Node* head)
{
    struct Node *slow_ptr = head,
                *fast_ptr = head;
    struct Node *second_half,
                *prev_of_slow_ptr = head;
 
    // To handle odd size list 
    struct Node* midnode = NULL;
    
    // Initialize result
    bool res = true;
 
    if (head != NULL &&
        head->next != NULL)
    {
        // Get the middle of the list.
        // Move slow_ptr by 1 and
        // fast_ptrr by 2, slow_ptr
        // will have the middle node
        while (fast_ptr != NULL &&
               fast_ptr->next != NULL)
        {
            fast_ptr = fast_ptr->next->next;
 
            // We need previous of the slow_ptr for
            // linked lists with odd elements
            prev_of_slow_ptr = slow_ptr;
            slow_ptr = slow_ptr->next;
        }
 
        /* fast_ptr would become NULL when there
           are even elements in list. And not NULL
           for odd elements. We need to skip the
           middle node for odd case and store it
           somewhere so that we can restore the
           original list*/
        if (fast_ptr != NULL)
        {
            midnode = slow_ptr;
            slow_ptr = slow_ptr->next;
        }
 
        // Now reverse the second half and
        // compare it with first half
        second_half = slow_ptr;
 
        // NULL terminate first half
        prev_of_slow_ptr->next = NULL;
 
        // Reverse the second half
        reverse(&second_half);
 
        // Compare
        res = compareLists(head, second_half);
 
        // Construct the original list back
        // Reverse the second half again
        reverse(&second_half);
 
        // If there was a mid node (odd size
        // case) which was not part of either
        // first half or second half.
        if (midnode != NULL)
        {
            prev_of_slow_ptr->next = midnode;
            midnode->next = second_half;
        }
        else
            prev_of_slow_ptr->next = second_half;
    }
    return res;
}
 
// Function to reverse the linked list
// Note that this function may change
// the head
void reverse(struct Node** head_ref)
{
    struct Node* prev = NULL;
    struct Node* current = *head_ref;
    struct Node* next;
    while (current != NULL)
    {
        next = current->next;
        current->next = prev;
        prev = current;
        current = next;
    }
    *head_ref = prev;
}
 
// Function to check if two input
// lists have same data
bool compareLists(struct Node* head1,
                  struct Node* head2)
{
    struct Node* temp1 = head1;
    struct Node* temp2 = head2;
 
    while (temp1 && temp2)
    {
        if (temp1->data == temp2->data)
        {
            temp1 = temp1->next;
            temp2 = temp2->next;
        }
        else
            return 0;
    }
 
    // Both are empty return 1
    if (temp1 == NULL && temp2 == NULL)
        return 1;
 
    // Will reach here when one is NULL
    // and other is not
    return 0;
}
 
// Push a node to linked list.
// Note that this function
// changes the head
void push(struct Node** head_ref,
          char new_data)
{
    // allocate node
    struct Node* new_node =
           (struct Node*)malloc(sizeof(struct Node));
 
    // Put in the data
    new_node->data = new_data;
 
    // Link the old list off the new node
    new_node->next = (*head_ref);
 
    // Move the head to point to the new node
    (*head_ref) = new_node;
}
 
// A utility function to print a
// given linked list
void printList(struct Node* ptr)
{
    while (ptr != NULL)
    {
        printf("%c->", ptr->data);
        ptr = ptr->next;
    }
    printf("NULL");
}
 
// Driver code
int main()
{
    // Start with the empty list
    struct Node* head = NULL;
    char str[] = "abacaba";
    int i;
 
    for (i = 0; str[i] != ''; i++)
    {
        push(&head, str[i]);
        printList(head);
        isPalindrome(head) ? printf("Is Palindrome") : printf("Not Palindrome");
    }
    return 0;
}

输出：

a->NULL
Is Palindrome

b->a->NULL
Not Palindrome

a->b->a->NULL
Is Palindrome

c->a->b->a->NULL
Not Palindrome

a->c->a->b->a->NULL
Not Palindrome

b->a->c->a->b->a->NULL
Not Palindrome

a->b->a->c->a->b->a->NULL
Is Palindrome

时间复杂度： O(n)
辅助空间： O(1)

方法 2（使用递归）：
使用左右两个指针。使用递归左右移动并检查每个递归调用中的跟随。
1）子列表是回文。
2）当前左右的值是匹配的。

如果上述两个条件都为真，则返回真。

这个想法是使用函数调用堆栈作为容器。递归遍历直到列表的末尾。当我们从最后一个 NULL 返回时，我们将在最后一个节点。最后一个节点将与列表的第一个节点进行比较。

为了访问列表的第一个节点，我们需要列表头在最后一次递归调用中可用。因此，我们也将 head 传递给递归函数。如果它们都匹配，我们需要比较 (2, n-2) 个节点。同样，当递归回退到 (n-2)nd 节点时，我们需要从头开始引用第二个节点。我们在前一个调用中推进头指针，以指向列表中的下一个节点。
然而，诀窍是识别双指针。传递单个指针与传递值一样好，我们将一次又一次地传递同一个指针。我们需要传递头指针的地址来反映父递归调用的变化。
感谢Sharad Chandra提出这种方法。

C

// Recursive program to check if a
// given linked list is palindrome
#include 
#include 
#include 
 
// Link list node
struct node
{
    char data;
    struct node* next;
};
 
// Initial parameters to this
// function are &head and head
bool isPalindromeUtil(struct node** left,
                      struct node* right)
{
    // Stop recursion when right
    // becomes NULL
    if (right == NULL)
        return true;
 
    // If sub-list is not palindrome then
    // no need to check for current left
    // and right, return false
    bool isp = isPalindromeUtil(left,
                                right->next);
    if (isp == false)
        return false;
 
    // Check values at current left and right
    bool isp1 = (right->data == (*left)->data);
 
    // Move left to next node
    *left = (*left)->next;
 
    return isp1;
}
 
// A wrapper over isPalindromeUtil()
bool isPalindrome(struct node* head)
{
    isPalindromeUtil(&head, head);
}
 
// Push a node to linked list.
// Note that this function changes
// the head
void push(struct node** head_ref,
          char new_data)
{
    // Allocate node
    struct node* new_node =
          (struct node*)malloc(sizeof(struct node));
 
    // Put in the data
    new_node->data = new_data;
 
    // Link the old list off the new node
    new_node->next = (*head_ref);
 
    // Move the head to point to the new node
    (*head_ref) = new_node;
}
 
// A utility function to print a
// given linked list
void printList(struct node* ptr)
{
    while (ptr != NULL)
    {
        printf("%c->", ptr->data);
        ptr = ptr->next;
    }
    printf("NULL\n");
}
 
// Driver code
int main()
{
    // Start with the empty list
    struct node* head = NULL;
    char str[] = "abacaba";
    int i;
 
    for (i = 0; str[i] != '\0'; i++)
    {
        push(&head, str[i]);
        printList(head);
        isPalindrome(head) ? printf("Is Palindrome\n\n") : printf("Not Palindrome\n\n");
    }
 
    return 0;
}

输出：

a->NULL
Not Palindrome

b->a->NULL
Not Palindrome

a->b->a->NULL
Is Palindrome

c->a->b->a->NULL
Not Palindrome

a->c->a->b->a->NULL
Not Palindrome

b->a->c->a->b->a->NULL
Not Palindrome

a->b->a->c->a->b->a->NULL
Is Palindrome

时间复杂度： O(n)
辅助空间：如果考虑函数调用堆栈大小，则为 O(n)，否则为 O(1)。

请参阅完整的函数文章来检查单链表是否为回文以获取更多详细信息！