# 算法设计：检查单链表是否为回文的函数

2021年3月11日18:15:04 发表评论 423 次浏览

## 推荐：请在"实践首先, 在继续解决方案之前。

• 一个简单的解决方案是使用列表节点堆栈。这主要涉及三个步骤。
• 从头到尾遍历给定列表, 并将每个访问的节点压入堆栈。
• 再次遍历列表。对于每个访问的节点, 从堆栈中弹出一个节点, 并将弹出的节点的数据与当前访问的节点进行比较。
• 如果所有节点都匹配, 则返回true, 否则返回false。

## C ++

``````#include<bits/stdc++.h>
using namespace std;

class Node {
public :
int data;
Node( int d){
data = d;
}
Node *ptr;
};

// Function to check if the linked list
// is palindrome or not

// Temp pointer

// Declare a stack
stack < int > s;

// Push all elements of the list
// to the stack
while (slow != NULL){
s.push(slow->data);

slow = slow->ptr;
}

// Iterate in the list again and
// check by popping from the stack

// Get the top most element
int i=s.top();

// Pop the element
s.pop();

// Check if data is not
// same as popped element
if (head -> data != i){
return false ;
}

}

return true ;
}

// Driver Code
int main(){

Node one =  Node(1);
Node two = Node(2);
Node three = Node(3);
Node four = Node(2);
Node five = Node(1);

// Initialize the next pointer
// of every current pointer
five.ptr = NULL;
one.ptr = &two;
two.ptr = &three;
three.ptr = &four;
four.ptr = &five;
Node* temp = &one;

// Call function to check palindrome or not
int result = isPalin(&one);

if (result == 1)
cout<< "isPalindrome is true\n" ;
else
cout<< "isPalindrome is true\n" ;

return 0;
}

// This code has been contributed by Striver``````

## Java

``````/* Java program to check if linked list is palindrome recursively */
import java.util.*;

public static void main(String args[])
{
Node one = new Node( 1 );
Node two = new Node( 2 );
Node three = new Node( 3 );
Node four = new Node( 4 );
Node five = new Node( 3 );
Node six = new Node( 2 );
Node seven = new Node( 1 );
one.ptr = two;
two.ptr = three;
three.ptr = four;
four.ptr = five;
five.ptr = six;
six.ptr = seven;
boolean condition = isPalindrome(one);
System.out.println( "isPalidrome :" + condition);
}
{

boolean ispalin = true ;
Stack<Integer> stack = new Stack<Integer>();

while (slow != null ) {
stack.push(slow.data);
slow = slow.ptr;
}

while (head != null ) {

int i = stack.pop();
ispalin = true ;
}
else {
ispalin = false ;
break ;
}
}
return ispalin;
}
}

class Node {
int data;
Node ptr;
Node( int d)
{
ptr = null ;
data = d;
}
}``````

## Python3

``````# Python3 program to check if linked
# list is palindrome using stack
class Node:
def __init__( self , data):

self .data = data
self .ptr = None

# Function to check if the linked list
# is palindrome or not

# Temp pointer

# Declare a stack
stack = []

ispalin = True

# Push all elements of the list
# to the stack
while slow ! = None :
stack.append(slow.data)

slow = slow.ptr

# Iterate in the list again and
# check by popping from the stack
while head ! = None :

# Get the top most element
i = stack.pop()

# Check if data is not
# same as popped element
ispalin = True
else :
ispalin = False
break

return ispalin

# Driver Code

one = Node( 1 )
two = Node( 2 )
three = Node( 3 )
four = Node( 4 )
five = Node( 3 )
six = Node( 2 )
seven = Node( 1 )

# Initialize the next pointer
# of every current pointer
one.ptr = two
two.ptr = three
three.ptr = four
four.ptr = five
five.ptr = six
six.ptr = seven
seven.ptr = None

# Call function to check palindrome or not
result = ispalindrome(one)

print ( "isPalindrome:" , result)

# This code is contributed by Nishtha Goel``````

## C#

``````// C# program to check if linked list
// is palindrome recursively
using System;
using System.Collections.Generic;

// Driver code
public static void Main(String []args)
{
Node one = new Node(1);
Node two = new Node(2);
Node three = new Node(3);
Node four = new Node(4);
Node five = new Node(3);
Node six = new Node(2);
Node seven = new Node(1);

one.ptr = two;
two.ptr = three;
three.ptr = four;
four.ptr = five;
five.ptr = six;
six.ptr = seven;

bool condition = isPalindrome(one);
Console.WriteLine( "isPalidrome :" + condition);
}

{
bool ispalin = true ;
Stack< int > stack = new Stack< int >();

while (slow != null )
{
stack.Push(slow.data);
slow = slow.ptr;
}

{
int i = stack.Pop();
{
ispalin = true ;
}
else
{
ispalin = false ;
break ;
}
}
return ispalin;
}
}

class Node
{
public int data;
public Node ptr;
public Node( int d)
{
ptr = null ;
data = d;
}
}

// This code is contributed by amal kumar choubey``````

``isPalindrome: true``

1)

2)

3)

4)

## C

``````/* Program to check if a linked list is palindrome */
#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>

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 */
{
struct Node *second_half, *prev_of_slow_ptr = head;
struct Node* midnode = NULL; // To handle odd size list
bool res = true ; // initialize result

/* 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;
prev_of_slow_ptr->next = NULL; // NULL terminate first half
reverse(&second_half); // Reverse the second half
res = compareLists(head, second_half); // compare

/* Construct the original list back */
reverse(&second_half); // Reverse the second half again

// 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 */
{
struct Node* prev = NULL;
struct Node* next;
while (current != NULL) {
next = current->next;
current->next = prev;
prev = current;
current = next;
}
}

/* Function to check if two input lists have same data*/
{

while (temp1 && temp2) {
if (temp1->data == temp2->data) {
temp1 = temp1->next;
temp2 = temp2->next;
}
else
return 0;
}

/* Both are empty reurn 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
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 */

/* move the head to pochar to the 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" );
}

/* Drier program to test above function*/
int main()
{
char str[] = "abacaba" ;
int i;

for (i = 0; str[i] != '\0' ; i++) {
isPalindrome(head) ? printf ( "Is Palindrome\n\n" ) : printf ( "Not Palindrome\n\n" );
}

return 0;
}``````

## Java

``````/* Java program to check if linked list is palindrome */

Node slow_ptr, fast_ptr, second_half;

class Node {
char data;
Node next;

Node( char d)
{
data = d;
next = null ;
}
}

/* Function to check if given linked list is
palindrome or not */
{
Node midnode = null ; // To handle odd size list
boolean res = true ; // initialize result

/* 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 the 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;
prev_of_slow_ptr.next = null ; // NULL terminate first half
reverse(); // Reverse the second half
res = compareLists(head, second_half); // compare

/* Construct the original list back */
reverse(); // Reverse the second half again

if (midnode != null ) {
// If there was a mid node (odd size case) which
// was not part of either first half or second half.
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()
{
Node prev = null ;
Node current = second_half;
Node next;
while (current != null ) {
next = current.next;
current.next = prev;
prev = current;
current = next;
}
second_half = prev;
}

/* Function to check if two input lists have same data*/
{

while (temp1 != null && temp2 != null ) {
if (temp1.data == temp2.data) {
temp1 = temp1.next;
temp2 = temp2.next;
}
else
return false ;
}

/* Both are empty reurn 1*/
if (temp1 == null && temp2 == null )
return true ;

/* Will reach here when one is NULL
and other is not */
return false ;
}

/* Push a node to linked list. Note that this function
public void push( char new_data)
{
/* Allocate the Node &
Put in the data */
Node new_node = new Node(new_data);

/* link the old list off the new one */

/* Move the head to point to new Node */
}

// A utility function to print a given linked list
void printList(Node ptr)
{
while (ptr != null ) {
System.out.print(ptr.data + "->" );
ptr = ptr.next;
}
System.out.println( "NULL" );
}

/* Driver program to test the above functions */
public static void main(String[] args)
{

char str[] = { 'a' , 'b' , 'a' , 'c' , 'a' , 'b' , 'a' };
String string = new String(str);
for ( int i = 0 ; i < 7 ; i++) {
llist.push(str[i]);
if (llist.isPalindrome(llist.head) != false ) {
System.out.println( "Is Palindrome" );
System.out.println( "" );
}
else {
System.out.println( "Not Palindrome" );
System.out.println( "" );
}
}
}
}``````

## C#

``````/* C# program to check if linked list is palindrome */
using System;
Node slow_ptr, fast_ptr, second_half;

public class Node {
public char data;
public Node next;

public Node( char d)
{
data = d;
next = null ;
}
}

/* Function to check if given linked list is
palindrome or not */
{
Node midnode = null ; // To handle odd size list
Boolean res = true ; // initialize result

/* 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 the 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;
prev_of_slow_ptr.next = null ; // NULL terminate first half
reverse(); // Reverse the second half
res = compareLists(head, second_half); // compare

/* Construct the original list back */
reverse(); // Reverse the second half again

if (midnode != null ) {
// If there was a mid node (odd size case) which
// was not part of either first half or second half.
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()
{
Node prev = null ;
Node current = second_half;
Node next;
while (current != null ) {
next = current.next;
current.next = prev;
prev = current;
current = next;
}
second_half = prev;
}

/* Function to check if two input lists have same data*/
{

while (temp1 != null && temp2 != null ) {
if (temp1.data == temp2.data) {
temp1 = temp1.next;
temp2 = temp2.next;
}
else
return false ;
}

/* Both are empty reurn 1*/
if (temp1 == null && temp2 == null )
return true ;

/* Will reach here when one is NULL
and other is not */
return false ;
}

/* Push a node to linked list. Note that this function
public void push( char new_data)
{
/* Allocate the Node &
Put in the data */
Node new_node = new Node(new_data);

/* link the old list off the new one */

/* Move the head to point to new Node */
}

// A utility function to print a given linked list
void printList(Node ptr)
{
while (ptr != null ) {
Console.Write(ptr.data + "->" );
ptr = ptr.next;
}
Console.WriteLine( "NULL" );
}

/* Driver program to test the above functions */
public static void Main(String[] args)
{

char [] str = { 'a' , 'b' , 'a' , 'c' , 'a' , 'b' , 'a' };

for ( int i = 0; i < 7; i++) {
llist.push(str[i]);
if (llist.isPalindrome(llist.head) != false ) {
Console.WriteLine( "Is Palindrome" );
Console.WriteLine( "" );
}
else {
Console.WriteLine( "Not Palindrome" );
Console.WriteLine( "" );
}
}
}
}
// This code is contributed by Arnab Kundu``````

``````a->NULL
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``````

O(1)

1)子列表是回文

2)当前左侧和右侧的值匹配。

## C ++

``````// Recursive program to check if a given linked list is palindrome
#include <bits/stdc++.h>
using namespace std;

struct node {
char data;
struct node* next;
};

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()
{
}

/* Push a node to linked list. Note that this function
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 */

/* move the head to pochar to the new node */
}

// A utility function to print a given linked list
void printList( struct node* ptr)
{
while (ptr != NULL) {
cout << ptr->data << "->" ;
ptr = ptr->next;
}
cout << "NULL\n" ;
}

/* Driver program to test above function*/
int main()
{
char str[] = "abacaba" ;
int i;

for (i = 0; str[i] != '\0' ; i++) {
isPalindrome(head) ? cout << "Is Palindrome\n\n" : cout << "Not Palindrome\n\n" ;
}

return 0;
}
//this code is contributed by shivanisinghss2110``````

## C

``````// Recursive program to check if a given linked list is palindrome
#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>

struct node {
char data;
struct node* next;
};

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()
{
}

/* Push a node to linked list. Note that this function
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 */

/* move the head to pochar to the 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 program to test above function*/
int main()
{
char str[] = "abacaba" ;
int i;

for (i = 0; str[i] != '\0' ; i++) {
isPalindrome(head) ? printf ( "Is Palindrome\n\n" ) : printf ( "Not Palindrome\n\n" );
}

return 0;
}``````

## Java

``````/* Java program to check if linked list is palindrome recursively */

Node left;

class Node {
char data;
Node next;

Node( char d)
{
data = d;
next = null ;
}
}

boolean isPalindromeUtil(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 */
boolean isp = isPalindromeUtil(right.next);
if (isp == false )
return false ;

/* Check values at current left and right */
boolean isp1 = (right.data == (left).data);

/* Move left to next node */
left = left.next;

return isp1;
}

// A wrapper over isPalindromeUtil()
{
return result;
}

/* Push a node to linked list. Note that this function
public void push( char new_data)
{
/* Allocate the Node &
Put in the data */
Node new_node = new Node(new_data);

/* link the old list off the new one */

/* Move the head to point to new Node */
}

// A utility function to print a given linked list
void printList(Node ptr)
{
while (ptr != null ) {
System.out.print(ptr.data + "->" );
ptr = ptr.next;
}
System.out.println( "NULL" );
}

/* Driver program to test the above functions */
public static void main(String[] args)
{

char str[] = { 'a' , 'b' , 'a' , 'c' , 'a' , 'b' , 'a' };
String string = new String(str);
for ( int i = 0 ; i < 7 ; i++) {
llist.push(str[i]);
if (llist.isPalindrome(llist.head) != false ) {
System.out.println( "Is Palindrome" );
System.out.println( "" );
}
else {
System.out.println( "Not Palindrome" );
System.out.println( "" );
}
}
}
}

// This code has been contributed by Mayank Jaiswal(mayank_24)``````

## C#

``````/* C# program to check if linked list
is palindrome recursively */
using System;

{
Node left;

public class Node
{
public char data;
public Node next;

public Node( char d)
{
data = d;
next = null ;
}
}

Boolean isPalindromeUtil(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 */
Boolean isp = isPalindromeUtil(right.next);
if (isp == false )
return false ;

/* Check values at current left and right */
Boolean isp1 = (right.data == (left).data);

/* Move left to next node */
left = left.next;

return isp1;
}

// A wrapper over isPalindromeUtil()
{
return result;
}

/* Push a node to linked list. Note that this function
public void push( char new_data)
{
/* Allocate the Node &
Put in the data */
Node new_node = new Node(new_data);

/* link the old list off the new one */

/* Move the head to point to new Node */
}

// A utility function to print a given linked list
void printList(Node ptr)
{
while (ptr != null )
{
Console.Write(ptr.data + "->" );
ptr = ptr.next;
}
Console.WriteLine( "NULL" );
}

/* Driver code */
public static void Main(String[] args)
{

char []str = { 'a' , 'b' , 'a' , 'c' , 'a' , 'b' , 'a' };
//String string = new String(str);
for ( int i = 0; i < 7; i++) {
llist.push(str[i]);
{
Console.WriteLine( "Is Palindrome" );
Console.WriteLine( "" );
}
else
{
Console.WriteLine( "Not Palindrome" );
Console.WriteLine( "" );
}
}
}
}

// This code is contributed by Rajput-Ji``````

``````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``````