将所有出现的元素移动到链表的结尾

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

一个简单的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末尾为止。

时间复杂度：O(n ² )

高效解决方案1：是要保留两个指针：
pCrawl =>指针，用于遍历整个列表。
pKey =>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。仅当pKey不指向键时，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之前的密钥，因此我们交换pCrawl和pKey的数据，并将pKey移到下一个位置。交换数据后，循环不变式是从pKey到pCrawl的所有元素都是键。

以下是此方法的实现。

C++

// C++ program to move all occurrences of a
// given key to end.
#include 
 
// 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

Python3

# Python3 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)仅需要遍历一个列表。

高效解决方案2：
1.遍历链表，然后将指针指向尾部。
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
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);
    }
}

Python3

# Python3 code to remove key element to
# end of linked list
  
# A Linked list Node
class Node:
     
    def __init__(self, data):
         
        self.data = data
        self.next = None
         
# A utility function to create a new node.
def newNode(x):
     
    temp = Node(x)
    return temp
 
# Function to remove key to end
def keyToEnd(head, key):
 
    # Node to keep pointing to tail
    tail = head
     
    if (head == None):
        return None
     
    while (tail.next != None):
        tail = tail.next
     
    # Node to point to last of linked list
    last = tail
    current = head
    prev = None
      
    # Node prev2 to point to previous
    # when head.data!=key
    prev2 = None
      
    # Loop to perform operations to
    # remove key to end
    while (current != tail):
        if (current.data == key and prev2 == None):
            prev = current
            current = current.next
            head = current
            last.next = prev
            last = last.next
            last.next = None
            prev = None
 
        else:
            if (current.data == key and prev2 != None):
                prev = current
                current = current.next
                prev2.next = current
                last.next = prev
                last = last.next
                last.next = None
             
            elif (current != tail):
                prev2 = current
                current = current.next
                 
    return head
 
# Function to display linked list
def printList(head):
 
    temp = head
     
    while (temp != None):
        print(temp.data, end = ' ')
        temp = temp.next
     
    print()
     
# Driver Code
if __name__=='__main__':
     
    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)
  
    key = 2
    print("Linked List before operations :")
    printList(root)
    print("Linked List after operations :")
    root = keyToEnd(root, key)
     
    printList(root)
     
# This code is contributed by rutvik_56

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。