编写一个C函数，将第一个元素移动到给定的单链表的末尾。例如，如果给定的链表是1->2->3->4->5，那么函数应该将列表更改为2->3->4->5->1。
null
算法：
遍历列表直到最后一个节点。使用两个指针：一个用于存储最后一个节点（last）的地址，另一个用于存储第一个节点（first）的地址。循环结束后，执行以下操作。 i） 将头部作为第二个节点（*head_ref=first->next）。 ii）将第一个的下一个设置为空（第一个->下一个=空）。 iii）将最后一个的下一个设置为第一个（最后一个->下一个=第一个）
C++
/* C++ Program to move first element to end in a given linked list */ #include <stdio.h> #include <stdlib.h> /* A linked list node */ struct Node { int data; struct Node* next; }; /* We are using a double pointer head_ref here because we change head of the linked list inside this function.*/ void moveToEnd( struct Node** head_ref) { /* If linked list is empty, or it contains only one node, then nothing needs to be done, simply return */ if (*head_ref == NULL || (*head_ref)->next == NULL) return ; /* Initialize first and last pointers */ struct Node* first = *head_ref; struct Node* last = *head_ref; /*After this loop last contains address of last node in Linked List */ while (last->next != NULL) { last = last->next; } /* Change the head pointer to point to second node now */ *head_ref = first->next; /* Set the next of first as NULL */ first->next = NULL; /* Set the next of last as first */ last->next = first; } /* UTILITY FUNCTIONS */ /* Function to add a node at the beginning of Linked List */ void push( struct Node** head_ref, int new_data) { struct Node* new_node = new Node; new_node->data = new_data; new_node->next = (*head_ref); (*head_ref) = new_node; } /* Function to print nodes in a given linked list */ void printList( struct Node* node) { while (node != NULL) { printf ( "%d " , node->data); node = node->next; } } /* Driver program to test above function */ int main() { struct Node* start = NULL; /* The constructed linked list is: 1->2->3->4->5 */ push(&start, 5); push(&start, 4); push(&start, 3); push(&start, 2); push(&start, 1); printf ( " Linked list before moving first to end" ); printList(start); moveToEnd(&start); printf ( " Linked list after moving first to end" ); printList(start); return 0; } |
JAVA
/* Java Program to move first element to end in a given linked list */ class Sol { /* A linked list node */ static class Node { int data; Node next; }; /* We are using a double pointer head_ref here because we change head of the linked list inside this function.*/ static Node moveToEnd( Node head_ref) { /* If linked list is empty, or it contains only one node, then nothing needs to be done, simply return */ if (head_ref == null || (head_ref).next == null ) return null ; /* Initialize first and last pointers */ Node first = head_ref; Node last = head_ref; /*After this loop last contains address of last node in Linked List */ while (last.next != null ) { last = last.next; } /* Change the head pointer to point to second node now */ head_ref = first.next; /* Set the next of first as null */ first.next = null ; /* Set the next of last as first */ last.next = first; return head_ref; } /* UTILITY FUNCTIONS */ /* Function to add a node at the beginning of Linked List */ static Node push( Node head_ref, int new_data) { Node new_node = new Node(); new_node.data = new_data; new_node.next = (head_ref); (head_ref) = new_node; return head_ref; } /* Function to print nodes in a given linked list */ static void printList( Node node) { while (node != null ) { System.out.printf( "%d " , node.data); node = node.next; } } /* Driver code */ public static void main(String args[]) { Node start = null ; /* The constructed linked list is: 1.2.3.4.5 */ start = push(start, 5 ); start = push(start, 4 ); start = push(start, 3 ); start = push(start, 2 ); start = push(start, 1 ); System.out.printf( " Linked list before moving first to end" ); printList(start); start = moveToEnd(start); System.out.printf( " Linked list after moving first to end" ); printList(start); } } // This code is contributed by Arnab Kundu |
Python3
# Python3 Program to move first element to end # in a given linked list # A linked list node class Node: def __init__( self ): self .data = 0 self . next = None # We are using a double pointer head_ref # here because we change head of the linked # list inside this function. def moveToEnd( head_ref) : # If linked list is empty, or it contains # only one node, then nothing needs to be # done, simply return if (head_ref = = None or (head_ref). next = = None ) : return None # Initialize first and last pointers first = head_ref last = head_ref # After this loop last contains address # of last node in Linked List while (last. next ! = None ) : last = last. next # Change the head pointer to point # to second node now head_ref = first. next # Set the next of first as None first. next = None # Set the next of last as first last. next = first return head_ref # UTILITY FUNCTIONS # Function to add a node at the beginning # of Linked List def push( head_ref, new_data) : new_node = Node() new_node.data = new_data new_node. next = (head_ref) (head_ref) = new_node return head_ref # Function to print nodes in a given linked list def printList(node) : while (node ! = None ): print (node.data, end = " " ) node = node. next # Driver code start = None # The constructed linked list is: #1.2.3.4.5 start = push(start, 5 ) start = push(start, 4 ) start = push(start, 3 ) start = push(start, 2 ) start = push(start, 1 ) print ( " Linked list before moving first to end" ) printList(start) start = moveToEnd(start) print ( " Linked list after moving first to end" ) printList(start) # This code is contributed by Arnab Kundu |
C#
/* C# Program to move first element to end in a given linked list */ using System; class GFG { /* A linked list node */ public class Node { public int data; public Node next; }; /* We are using a double pointer head_ref here because we change head of the linked list inside this function.*/ static Node moveToEnd( Node head_ref) { /* If linked list is empty, or it contains only one node, then nothing needs to be done, simply return */ if (head_ref == null || (head_ref).next == null ) return null ; /* Initialize first and last pointers */ Node first = head_ref; Node last = head_ref; /*After this loop last contains address of last node in Linked List */ while (last.next != null ) { last = last.next; } /* Change the head pointer to point to second node now */ head_ref = first.next; /* Set the next of first as null */ first.next = null ; /* Set the next of last as first */ last.next = first; return head_ref; } /* UTILITY FUNCTIONS */ /* Function to add a node at the beginning of Linked List */ static Node push( Node head_ref, int new_data) { Node new_node = new Node(); new_node.data = new_data; new_node.next = (head_ref); (head_ref) = new_node; return head_ref; } /* Function to print nodes in a given linked list */ static void printList( Node node) { while (node != null ) { Console.Write( "{0} " , node.data); node = node.next; } } /* Driver code */ public static void Main(String []args) { Node start = null ; /* The constructed linked list is: 1.2.3.4.5 */ start = push(start, 5); start = push(start, 4); start = push(start, 3); start = push(start, 2); start = push(start, 1); Console.Write( " Linked list before moving first to end" ); printList(start); start = moveToEnd(start); Console.Write( " Linked list after moving first to end" ); printList(start); } } // This code is contributed by 29AjayKumar |
Javascript
<script> // JavaScript program to move first element // to end in a given linked list /* A linked list node */ class Node { constructor() { this .data = 0; this .next = null ; } } /* We are using a double pointer head_ref here because we change head of the linked list inside this function.*/ function moveToEnd(head_ref) { /* If linked list is empty, or it contains only one node, then nothing needs to be done, simply return */ if (head_ref == null || head_ref.next == null ) return null ; /* Initialize first and last pointers */ var first = head_ref; var last = head_ref; /*After this loop last contains address of last node in Linked List */ while (last.next != null ) { last = last.next; } /* Change the head pointer to point to second node now */ head_ref = first.next; /* Set the next of first as null */ first.next = null ; /* Set the next of last as first */ last.next = first; return head_ref; } /* UTILITY FUNCTIONS */ /* Function to add a node at the beginning of Linked List */ function push(head_ref, new_data) { var new_node = new Node(); new_node.data = new_data; new_node.next = head_ref; head_ref = new_node; return head_ref; } // Function to print nodes in // a given linked list function printList(node) { while (node != null ) { document.write(node.data + " " ); node = node.next; } } // Driver code var start = null ; // The constructed linked list is: // 1.2.3.4.5 start = push(start, 5); start = push(start, 4); start = push(start, 3); start = push(start, 2); start = push(start, 1); document.write( "Linked list before " + "moving first to end<br>" ); printList(start); start = moveToEnd(start); document.write( "<br> Linked list after moving " + "first to end<br>" ); printList(start); // This code is contributed by rdtank </script> |
输出：
Linked list before moving first to end1 2 3 4 5 Linked list after moving first to end2 3 4 5 1
时间复杂性： O（n），其中n是给定链表中的节点数。
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