本文概述
- 建议：在继续解决方案之前, 请先在{IDE}上尝试使用你的方法。
- C ++
- Java
- python
- C#
- C ++
- Java
- C#
【算法（将所有出现的元素移动到链表的结尾）】例子：
Input: 1 ->
2 ->
2 ->
4 ->
3key = 2 Output : 1 ->
4 ->
3 ->
2 ->
2Input: 6 ->
6 ->
7 ->
6 ->
3 ->
10key = 6Output : 7 ->
3 ->
10 ->
6 ->
6 ->
6推荐：请尝试以下方法{IDE}首先, 在继续解决方案之前。一种简单的解决方案一张一张地找到链表中给定密钥的所有出现。对于每个发现的事件, 将其插入末尾。直到所有出现的给定键都移到结尾为止。
时间复杂度：O(n2)
高效解决方案1：
是保持两个指针：
爬行
=> 用于遍历整个列表的指针。
键
=> 如果找到了密钥, 则指向发生密钥的指针。其他与pCrawl相同。
我们从链接列表的开头开始上述两个指针。我们走键只有当键没有指向钥匙。我们总是搬家爬行。所以什么时候爬行和键是不一样的, 我们必须找到一个位于爬行, 因此我们交换的数据爬行和键, 然后移动键到下一个位置。交换数据后, 循环不变式是来自键to爬行是钥匙。
下面是此方法的实现。
C ++
// C++ program to move all occurrences of a
// given key to end.
#include <
bits/stdc++.h>
// A Linked list Node
struct Node {
int data;
struct Node* next;
};
// A urility function to create a new node.
struct Node* newNode( int x)
{
Node* temp = new Node;
temp->
data = https://www.lsbin.com/x;
temp->
next = NULL;
}// Utility function to print the elements
// in Linked list
void printList(Node* head)
{
struct Node* temp = head;
while (temp != NULL) {
printf ("%d " , temp->
data);
temp = temp->
next;
}
printf ( "\n" );
}// Moves all occurrences of given key to
// end of linked list.
void moveToEnd(Node* head, int key)
{
// Keeps track of locations where key
// is present.
struct Node* pKey = head;
// Traverse list
struct Node* pCrawl = head;
while (pCrawl != NULL) {
// If current pointer is not same as pointer
// to a key location, then we must have found
// a key in linked list. We swap data of pCrawl
// and pKey and move pKey to next position.
if (pCrawl != pKey &
&
pCrawl->
data != key) {
pKey->
data = https://www.lsbin.com/pCrawl->
data;
pCrawl->
data = key;
pKey = pKey->
next;
}// Find next position where key is present
if (pKey->
data != key)
pKey = pKey->
next;
// Moving to next Node
pCrawl = pCrawl->
next;
}
}// Driver code
int main()
{
Node* head = newNode(10);
head->
next = newNode(20);
head->
next->
next = newNode(10);
head->
next->
next->
next = newNode(30);
head->
next->
next->
next->
next = newNode(40);
head->
next->
next->
next->
next->
next = newNode(10);
head->
next->
next->
next->
next->
next->
next = newNode(60);
printf ("Before moveToEnd(), the Linked list is\n" );
printList(head);
int key = 10;
moveToEnd(head, key);
printf ( "\nAfter moveToEnd(), the Linked list is\n" );
printList(head);
return 0;
}Java
// Java program to move all occurrences of a
// given key to end.
class GFG {// A Linked list Node
static class Node {
int data;
Node next;
}// A urility function to create a new node.
static Node newNode( int x)
{
Node temp = new Node();
temp.data = https://www.lsbin.com/x;
temp.next = null ;
return temp;
}// Utility function to print the elements
// in Linked list
static void printList(Node head)
{
Node temp = head;
while (temp != null ) {
System.out.printf("%d " , temp.data);
temp = temp.next;
}
System.out.printf( "\n" );
}// Moves all occurrences of given key to
// end of linked list.
static void moveToEnd(Node head, int key)
{
// Keeps track of locations where key
// is present.
Node pKey = head;
// Traverse list
Node pCrawl = head;
while (pCrawl != null ) {
// If current pointer is not same as pointer
// to a key location, then we must have found
// a key in linked list. We swap data of pCrawl
// and pKey and move pKey to next position.
if (pCrawl != pKey &
&
pCrawl.data != key) {
pKey.data = https://www.lsbin.com/pCrawl.data;
pCrawl.data = key;
pKey = pKey.next;
}// Find next position where key is present
if (pKey.data != key)
pKey = pKey.next;
// Moving to next Node
pCrawl = pCrawl.next;
}
}// Driver code
public static void main(String args[])
{
Node head = newNode( 10 );
head.next = newNode( 20 );
head.next.next = newNode( 10 );
head.next.next.next = newNode( 30 );
head.next.next.next.next = newNode( 40 );
head.next.next.next.next.next = newNode( 10 );
head.next.next.next.next.next.next = newNode( 60 );
System.out.printf("Before moveToEnd(), the Linked list is\n" );
printList(head);
int key = 10 ;
moveToEnd(head, key);
System.out.printf( "\nAfter moveToEnd(), the Linked list is\n" );
printList(head);
}
}// This code is contributed by Arnab Kundupython
# Python program to move all occurrences of a
# given key to end.# Linked List node
class Node:
def __init__( self , data):
self .data = https://www.lsbin.com/data
self . next = None# A urility function to create a new node.
def newNode(x):temp = Node( 0 )
temp.data = x
temp. next = None
return temp# Utility function to print the elements
# in Linked list
def printList( head):temp = head
while (temp ! = None ) :
print ( temp.data, end =" " )
temp = temp. nextprint ()# Moves all occurrences of given key to
# end of linked list.
def moveToEnd(head, key):# Keeps track of locations where key
# is present.
pKey = head# Traverse list
pCrawl = head
while (pCrawl ! = None ) :# If current pointer is not same as pointer
# to a key location, then we must have found
# a key in linked list. We swap data of pCrawl
# and pKey and move pKey to next position.
if (pCrawl ! = pKey and pCrawl.data ! = key) :
pKey.data = https://www.lsbin.com/pCrawl.data
pCrawl.data = key
pKey = pKey. next# Find next position where key is present
if (pKey.data ! = key):
pKey = pKey. next# Moving to next Node
pCrawl = pCrawl. nextreturn head# Driver code
head = newNode( 10 )
head. next = newNode( 20 )
head. next . next = newNode( 10 )
head. next . next . next = newNode( 30 )
head. next . next . next . next = newNode( 40 )
head. next . next . next . next . next = newNode( 10 )
head. next . next . next . next . next . next = newNode( 60 )print ("Before moveToEnd(), the Linked list is\n" )
printList(head)key = 10
head = moveToEnd(head, key)print ( "\nAfter moveToEnd(), the Linked list is\n" )
printList(head)# This code is contributed by Arnab KunduC#
// C# program to move all occurrences of a
// given key to end.
using System;
class GFG {// A Linked list Node
public class Node {
public int data;
public Node next;
}// A urility function to create a new node.
static Node newNode( int x)
{
Node temp = new Node();
temp.data = https://www.lsbin.com/x;
temp.next = null ;
return temp;
}// Utility function to print the elements
// in Linked list
static void printList(Node head)
{
Node temp = head;
while (temp != null ) {
Console.Write("{0} " , temp.data);
temp = temp.next;
}
Console.Write( "\n" );
}// Moves all occurrences of given key to
// end of linked list.
static void moveToEnd(Node head, int key)
{
// Keeps track of locations where key
// is present.
Node pKey = head;
// Traverse list
Node pCrawl = head;
while (pCrawl != null ) {
// If current pointer is not same as pointer
// to a key location, then we must have found
// a key in linked list. We swap data of pCrawl
// and pKey and move pKey to next position.
if (pCrawl != pKey &
&
pCrawl.data != key) {
pKey.data = https://www.lsbin.com/pCrawl.data;
pCrawl.data = key;
pKey = pKey.next;
}// Find next position where key is present
if (pKey.data != key)
pKey = pKey.next;
// Moving to next Node
pCrawl = pCrawl.next;
}
}// Driver code
public static void Main(String[] args)
{
Node head = newNode(10);
head.next = newNode(20);
head.next.next = newNode(10);
head.next.next.next = newNode(30);
head.next.next.next.next = newNode(40);
head.next.next.next.next.next = newNode(10);
head.next.next.next.next.next.next = newNode(60);
Console.Write("Before moveToEnd(), the Linked list is\n" );
printList(head);
int key = 10;
moveToEnd(head, key);
Console.Write( "\nAfter moveToEnd(), the Linked list is\n" );
printList(head);
}
}// This code has been contributed by 29AjayKumar输出如下：
Before moveToEnd(), the Linked list is10 20 10 30 40 10 60 After moveToEnd(), the Linked list is20 30 40 60 10 10 10时间复杂度：O(n)仅需要遍历list。
高效解决方案2：
1.遍历链接列表, 并在末尾指向一个指针。
2.现在, 检查密钥和node-> data, 如果它们相等, 则将节点移至last-next, 否则移至
先。
C ++
// C++ code to remove key element to end of linked list
#include<
bits/stdc++.h>
using namespace std;
// A Linked list Node
struct Node
{
int data;
struct Node* next;
};
// A urility function to create a new node.
struct Node* newNode( int x)
{
Node* temp = new Node;
temp->
data = https://www.lsbin.com/x;
temp->
next = NULL;
}// Function to remove key to end
Node *keyToEnd(Node* head, int key)
{// Node to keep pointing to tail
Node* tail = head;
if (head == NULL)
{
return NULL;
}
while (tail->
next != NULL)
{
tail = tail->
next;
}// Node to point to last of linked list
Node* last = tail;
Node* current = head;
Node* prev = NULL;
// Node prev2 to point to previous when head.data!=key
Node* prev2 = NULL;
// loop to perform operations to remove key to end
while (current != tail)
{
if (current->
data == key &
&
prev2 == NULL)
{
prev = current;
current = current->
next;
head = current;
last->
next = prev;
last = last->
next;
last->
next = NULL;
prev = NULL;
}
else
{
if (current->
data == key &
&
prev2 != NULL)
{
prev = current;
current = current->
next;
prev2->
next = current;
last->
next = prev;
last = last->
next;
last->
next = NULL;
}
else if (current != tail)
{
prev2 = current;
current = current->
next;
}
}
}
return head;
}// Function to display linked list
void printList(Node* head)
{
struct Node* temp = head;
while (temp != NULL)
{
printf ("%d " , temp->
data);
temp = temp->
next;
}
printf ( "\n" );
}// Driver Code
int main()
{
Node* root = newNode(5);
root->
next = newNode(2);
root->
next->
next = newNode(2);
root->
next->
next->
next = newNode(7);
root->
next->
next->
next->
next = newNode(2);
root->
next->
next->
next->
next->
next = newNode(2);
root->
next->
next->
next->
next->
next->
next = newNode(2);
int key = 2;
cout <
<
"Linked List before operations :" ;
printList(root);
cout <
<
"\nLinked List after operations :" ;
root = keyToEnd(root, key);
printList(root);
return 0;
}// This code is contributed by Rajout-JiJava
// Java code to remove key element to end of linked list
import java.util.*;
// Node class
class Node {
int data;
Node next;
public Node( int data)
{
this .data = https://www.lsbin.com/data;
this .next = null ;
}
}class gfg {static Node root;
// Function to remove key to end
public static Node keyToEnd(Node head, int key)
{// Node to keep pointing to tail
Node tail = head;
if (head == null ) {
return null ;
}while (tail.next != null ) {
tail = tail.next;
}// Node to point to last of linked list
Node last = tail;
Node current = head;
Node prev = null ;
// Node prev2 to point to previous when head.data!=key
Node prev2 = null ;
// loop to perform operations to remove key to end
while (current != tail) {
if (current.data == key &
&
prev2 == null ) {
prev = current;
current = current.next;
head = current;
last.next = prev;
last = last.next;
last.next = null ;
prev = null ;
}
else {
if (current.data == key &
&
prev2 != null ) {
prev = current;
current = current.next;
prev2.next = current;
last.next = prev;
last = last.next;
last.next = null ;
}
else if (current != tail) {
prev2 = current;
current = current.next;
}
}
}
return head;
}// Function to display linked list
public static void display(Node root)
{
while (root != null ) {
System.out.print(root.data +" " );
root = root.next;
}
}// Driver Code
public static void main(String args[])
{
root = new Node( 5 );
root.next = new Node( 2 );
root.next.next = new Node( 2 );
root.next.next.next = new Node( 7 );
root.next.next.next.next = new Node( 2 );
root.next.next.next.next.next = new Node( 2 );
root.next.next.next.next.next.next = new Node( 2 );
int key = 2 ;
System.out.println( "Linked List before operations :" );
display(root);
System.out.println( "\nLinked List after operations :" );
root = keyToEnd(root, key);
display(root);
}
}C#
// C# code to remove key
// element to end of linked list
using System;
// Node class
public class Node {
public int data;
public Node next;
public Node( int data)
{
this .data = https://www.lsbin.com/data;
this .next = null ;
}
}class GFG {static Node root;
// Function to remove key to end
public static Node keyToEnd(Node head, int key)
{// Node to keep pointing to tail
Node tail = head;
if (head == null ) {
return null ;
}while (tail.next != null ) {
tail = tail.next;
}// Node to point to last of linked list
Node last = tail;
Node current = head;
Node prev = null ;
// Node prev2 to point to
// previous when head.data!=key
Node prev2 = null ;
// loop to perform operations
// to remove key to end
while (current != tail) {
if (current.data == key &
&
prev2 == null ) {
prev = current;
current = current.next;
head = current;
last.next = prev;
last = last.next;
last.next = null ;
prev = null ;
}
else {
if (current.data == key &
&
prev2 != null ) {
prev = current;
current = current.next;
prev2.next = current;
last.next = prev;
last = last.next;
last.next = null ;
}
else if (current != tail) {
prev2 = current;
current = current.next;
}
}
}
return head;
}// Function to display linked list
public static void display(Node root)
{
while (root != null ) {
Console.Write(root.data +" " );
root = root.next;
}
}// Driver Code
public static void Main()
{
root = new Node(5);
root.next = new Node(2);
root.next.next = new Node(2);
root.next.next.next = new Node(7);
root.next.next.next.next = new Node(2);
root.next.next.next.next.next = new Node(2);
root.next.next.next.next.next.next = new Node(2);
int key = 2;
Console.WriteLine( "Linked List before operations :" );
display(root);
Console.WriteLine( "\nLinked List after operations :" );
root = keyToEnd(root, key);
display(root);
}
}// This code is contributed by PrinciRaj1992输出如下：
Linked List before operations :5 2 2 7 2 2 2 Linked List after operations :5 7 2 2 2 2 2谢谢拉文德·库马尔建议这种方法。
高效解决方案3：是维护一个单独的密钥列表。我们将此键列表初始化为空。我们遍历给定的列表。对于找到的每个密钥, 我们将其从原始列表中删除, 然后插入单独的密钥列表中。最后, 我们在剩余给定列表的末尾链接关键字列表。该解决方案的时间复杂度也是O(n), 并且它只需要遍历list。
如果发现任何不正确的地方, 或者想分享有关上述主题的更多信息, 请写评论。
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