C++ Program For Moving All Occurrences Of An Element To End In A Linked List
Last Updated :
23 Jul, 2025
Given a linked list and a key in it, the task is to move all occurrences of the given key to the end of the linked list, keeping the order of all other elements the same.
Examples:
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
A simple solution is to one by one find all occurrences of a given key in the linked list. For every found occurrence, insert it at the end. We do it till all occurrences of the given key are moved to the end.
Time Complexity: O(n2)
Efficient Solution 1: is to keep two pointers:
pCrawl => Pointer to traverse the whole list one by one.
pKey => Pointer to an occurrence of the key if a key is found. Else same as pCrawl.
We start both of the above pointers from the head of the linked list. We move pKey only when pKey is not pointing to a key. We always move pCrawl. So, when pCrawl and pKey are not the same, we must have found a key that lies before pCrawl, so we swap between pCrawl and pKey, and move pKey to the next location. The loop invariant is, after swapping of data, all elements from pKey to pCrawl are keys.
Below is the implementation of this approach.
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 utility 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("
");
}
// 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
");
printList(head);
int key = 10;
moveToEnd(head, key);
printf("
After moveToEnd(), the Linked list is
");
printList(head);
return 0;
}
Output:
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
Time Complexity: O(n) requires only one traversal of the list.
Auxiliary Space: O(1) as no extra space is required so it is a constant.
Efficient Solution 2 :
1. Traverse the linked list and take a pointer at the tail.
2. Now, check for the key and node->data. If they are equal, move the node to last-next, else move ahead.
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 utility 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
Output:
Linked List before operations :
5 2 2 7 2 2 2
Linked List after operations :
5 7 2 2 2 2 2
Time Complexity: O(n), where n represents the size of the given list.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
Thanks to Ravinder Kumar for suggesting this method.
Efficient Solution 3: is to maintain a separate list of keys. We initialize this list of keys as empty. We traverse the given list. For every key found, we remove it from the original list and insert it into a separate list of keys. We finally link the list of keys at the end of the remaining given list. The time complexity of this solution is also O(n) and it also requires only one traversal of the list.
Please refer complete article on Move all occurrences of an element to end in a linked list for more details!
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