本文概述
给定一个单字符列表, 编写一个函数, 如果给定列表是回文, 则返回true, 否则返回false。
推荐：请在"实践首先, 在继续解决方案之前。
方法1(使用堆栈)
- 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。
下图是上述方法的模拟：
下面是上述方法的实现：
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
bool isPalin(Node* head){
// Temp pointer
Node* slow= head;
// Declare a stack
stack < int > s;
// Push all elements of the list
// to the stack
while (slow != NULL){
s.push(slow->data);
// Move ahead
slow = slow->ptr;
}
// Iterate in the list again and
// check by popping from the stack
while (head != NULL ){
// 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 ;
}
// Move ahead
head=head->ptr;
}
return true ;
}
// Driver Code
int main(){
// Addition of linked list
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 StriverJava
/* Java program to check if linked list is palindrome recursively */
import java.util.*;
class linkeList {
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);
}
static boolean isPalindrome(Node head)
{
Node slow = head;
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();
if (head.data == i) {
ispalin = true ;
}
else {
ispalin = false ;
break ;
}
head = head.ptr;
}
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
def ispalindrome(head):
# Temp pointer
slow = head
# Declare a stack
stack = []
ispalin = True
# Push all elements of the list
# to the stack
while slow ! = None :
stack.append(slow.data)
# Move ahead
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
if head.data = = i:
ispalin = True
else :
ispalin = False
break
# Move ahead
head = head.ptr
return ispalin
# Driver Code
# Addition of linked list
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 GoelC#
// C# program to check if linked list
// is palindrome recursively
using System;
using System.Collections.Generic;
class linkeList{
// 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);
}
static bool isPalindrome(Node head)
{
Node slow = head;
bool ispalin = true ;
Stack< int > stack = new Stack< int >();
while (slow != null )
{
stack.Push(slow.data);
slow = slow.ptr;
}
while (head != null )
{
int i = stack.Pop();
if (head.data == i)
{
ispalin = true ;
}
else
{
ispalin = false ;
break ;
}
head = head.ptr;
}
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输出如下
isPalindrome: true上述方法的时间复杂度为O(n)。
方法2(通过反转列表)
此方法需要O(n)时间和O(1)额外空间。
1)
获取链接列表的中间位置。
2)
颠倒链接列表的后半部分。
3)
检查上半部分和下半部分是否相同。
4)
通过再次反转后半部分并将其附加到前半部分来构造原始链表
将列表分为两半, 方法2
这个
使用帖子。
当节点数为偶数时, 前一半和后一半恰好包含一半节点。这种方法的挑战性在于处理节点数为奇数的情况。我们不希望中间节点成为任何列表的一部分, 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" midnode"。
C
/* Program to check if a linked list is palindrome */
#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>
/* 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;
struct Node* midnode = NULL; // To handle odd size list
bool res = true ; // initialize result
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;
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 */
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 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
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 pochar 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" );
}
/* Drier program to test above function*/
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;
}Java
/* Java program to check if linked list is palindrome */
class LinkedList {
Node head; // head of list
Node slow_ptr, fast_ptr, second_half;
/* Linked list Node*/
class Node {
char data;
Node next;
Node( char d)
{
data = d;
next = null ;
}
}
/* Function to check if given linked list is
palindrome or not */
boolean isPalindrome(Node head)
{
slow_ptr = head;
fast_ptr = head;
Node prev_of_slow_ptr = head;
Node midnode = null ; // To handle odd size list
boolean res = true ; // initialize result
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 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*/
boolean compareLists(Node head1, Node head2)
{
Node temp1 = head1;
Node temp2 = head2;
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
changes the head */
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 */
new_node.next = head;
/* Move the head to point to new Node */
head = 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)
{
/* Start with the empty list */
LinkedList llist = new LinkedList();
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]);
llist.printList(llist.head);
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;
class LinkedList {
Node head; // head of list
Node slow_ptr, fast_ptr, second_half;
/* Linked list Node*/
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 */
Boolean isPalindrome(Node head)
{
slow_ptr = head;
fast_ptr = head;
Node prev_of_slow_ptr = head;
Node midnode = null ; // To handle odd size list
Boolean res = true ; // initialize result
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 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*/
Boolean compareLists(Node head1, Node head2)
{
Node temp1 = head1;
Node temp2 = head2;
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
changes the head */
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 */
new_node.next = head;
/* Move the head to point to new Node */
head = 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)
{
/* Start with the empty list */
LinkedList llist = new LinkedList();
char [] str = { 'a' , 'b' , 'a' , 'c' , 'a' , 'b' , 'a' };
for ( int i = 0; i < 7; i++) {
llist.push(str[i]);
llist.printList(llist.head);
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输出如下：
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时间复杂度：
上)
辅助空间：
O(1)
方法3(使用递归)
左右使用两个指针。使用递归在左右移动, 并在每个递归调用中检查是否跟随。
1)子列表是回文。
2)当前左侧和右侧的值匹配。
如果以上两个条件均成立, 则返回true。
这个想法是使用函数调用堆栈作为容器。递归遍历直到列表末尾。当我们从最后一个NULL返回时, 我们将在最后一个节点。要与列表的第一个节点进行比较的最后一个节点。
为了访问列表的第一个节点, 我们需要列表头在递归的最后一次调用中可用。因此, 我们也将head传递给了递归函数。如果它们都匹配, 则需要比较(2, n-2)个节点。同样, 当递归回落到第(n-2)个节点时, 我们需要从头开始引用第二个节点。我们在上一个调用中前进头指针, 以引用列表中的下一个节点。
然而, 识别双指针的技巧。传递单个指针与传递值一样好, 我们将一次又一次传递相同的指针。我们需要传递头部指针的地址以反映父递归调用中的更改。
谢谢
莎拉德·钱德拉(Sharad Chandra)
建议这种方法。
C ++
// Recursive program to check if a given linked list is palindrome
#include <bits/stdc++.h>
using namespace std;
/* 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 pochar 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) {
cout << ptr->data << "->" ;
ptr = ptr->next;
}
cout << "NULL\n" ;
}
/* Driver program to test above function*/
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) ? cout << "Is Palindrome\n\n" : cout << "Not Palindrome\n\n" ;
}
return 0;
}
//this code is contributed by shivanisinghss2110C
// Recursive program to check if a given linked list is palindrome
#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>
/* 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 pochar 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 program to test above function*/
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;
}Java
/* Java program to check if linked list is palindrome recursively */
class LinkedList {
Node head; // head of list
Node left;
/* Linked list Node*/
class Node {
char data;
Node next;
Node( char d)
{
data = d;
next = null ;
}
}
// Initial parameters to this function are &head and head
boolean isPalindromeUtil(Node right)
{
left = head;
/* 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()
boolean isPalindrome(Node head)
{
boolean result = isPalindromeUtil(head);
return result;
}
/* Push a node to linked list. Note that this function
changes the head */
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 */
new_node.next = head;
/* Move the head to point to new Node */
head = 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)
{
/* Start with the empty list */
LinkedList llist = new LinkedList();
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]);
llist.printList(llist.head);
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;
public class LinkedList
{
Node head; // head of list
Node left;
/* Linked list Node*/
public class Node
{
public char data;
public Node next;
public Node( char d)
{
data = d;
next = null ;
}
}
// Initial parameters to this function are &head and head
Boolean isPalindromeUtil(Node right)
{
left = head;
/* 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()
Boolean isPalindrome(Node head)
{
Boolean result = isPalindromeUtil(head);
return result;
}
/* Push a node to linked list. Note that this function
changes the head */
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 */
new_node.next = head;
/* Move the head to point to new Node */
head = 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)
{
/* Start with the empty list */
LinkedList llist = new LinkedList();
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]);
llist.printList(llist.head);
if (llist.isPalindrome(llist.head) != false )
{
Console.WriteLine( "Is Palindrome" );
Console.WriteLine( "" );
}
else
{
Console.WriteLine( "Not Palindrome" );
Console.WriteLine( "" );
}
}
}
}
// This code is contributed by Rajput-Ji输出如下：
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：
上)
辅助空间：
如果考虑函数调用堆栈的大小, 则为O(n), 否则为O(1)。
![从字法上最小长度N的排列，使得对于正好为K个索引，a[i] a[i]+1](https://www.lsbin.com/wp-content/themes/begin%20lts/img/loading.png)
