将所有出现的元素移到链表末尾
原文：https://www.geeksforgeeks.org/move-occurrences-element-end-linked-list/
给定一个链表和链表中的一个键，任务是将所有出现的给定键移动到链表的末尾，同时保持所有其他元素的顺序相同。
例子：
Input : 1 -> 2 -> 2 -> 4 -> 3
key = 2
Output : 1 -> 4 -> 3 -> 2 -> 2
Input : 6 -> 6 -> 7 -> 6 -> 3 -> 10
key = 6
Output : 7 -> 3 -> 10 -> 6 -> 6 -> 6
一个简单解决方案是一个一个地找到链表中所有给定键的出现。 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(N ^ 2)
高效解决方案 1：保留两个指针：
pCrawl：用来遍历整个列表的指针。
pKey：如果找到了键，则指向发生键的指针。 其他与pCrawl相同。
我们从链表的开头开始上述两个指针。 仅当pKey未指向键时，才移动pKey。 我们总是移动pCrawl。 因此，当pCrawl和pKey不相同时，我们必须找到一个位于pCrawl之前的键，因此我们交换pCrawl和pKey的数据，然后将pKey移动到下一个位置。 循环不变是交换数据后，从pKey到pCrawl的所有元素都是键。
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 = 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 = 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 = 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 = 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 Kundu
Python
# Python program to move all occurrences of a
# given key to end.
# Linked List node
class Node:
def __init__(self, data):
self.data = 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.next
print()
# 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 = 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
return 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 Kundu
C
// 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 = 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 = 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：
Before moveToEnd(), the Linked list is
10 20 10 30 40 10 60
After moveToEnd(), the Linked list is
20 30 40 60 10 10 10
时间复杂度：O(n)仅需要遍历列表。
有效解决方案 2：
-
遍历链表，并在末尾使用指针。
-
现在，检查键和节点->数据是否相等，将节点移至倒数第二个，否则向前移
。
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 = 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-Ji
Java
// 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 = 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 = 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：
Linked List before operations :
5 2 2 7 2 2 2
Linked List after operations :
5 7 2 2 2 2 2
感谢 Ravinder Kumar 提出了此方法。
高效解决方案 3：将维护单独的键列表。 我们将此键列表初始化为空。 我们遍历给定的列表。 对于找到的每个键，我们将其从原始列表中删除，然后插入单独的键列表中。 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)，并且它只需要遍历列表。
本文由 MAZHAR IMAM KHAN 撰写。 如果您喜欢 GeeksforGeeks 并希望做出贡献，则还可以使用 tribution.geeksforgeeks.org 撰写文章，或将您的文章邮寄至 tribution@geeksforgeeks.org。 查看您的文章出现在 GeeksforGeeks 主页上，并帮助其他 Geeks。
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