Given a linked list and a key in it, the task is to move all occurrences of given key to end of linked list, keeping order of all other elements 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 given key in linked list. For every found occurrence, insert it at the end. We do it till all occurrences of given key are moved to end.

Time Complexity : O(n^{2})

An **Efficient Solution** is to keep two pointers:

**pCrawl** => Pointer to traverse the whole list one by one.

**pKey** => Pointer to an occurrence of key if a key is found. Else same as pCrawl.

We start both of the above pointers from head of 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 same, we must have found a key which lies before **pCrawl**, so we swap data of **pCrawl** and **pKey**, and move **pKey** to next location. The loop invariant is, after swapping of data, all elements from **pKey** to **pCrawl** are keys.

Below is the C++ implementation of this approach.

// 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; }

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 list.

**Another Efficient Solution** is to maintain a separate list of keys. We initialize this list of keys as empty. We traverse given list. For every key found, we remove it from the original list and insert into the separate list of keys. We finally link list of keys at the end of remaining given list. Time complexity of this solution is also O(n) and it also requires only one traversal of list.

This article is contributed by **MAZHAR IMAM KHAN**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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