Given a linked list and a value x, partition a linked list around a value x, such that all nodes less than x come before all nodes greater than or equal to x. If x is contained within the list the values of x only need to be after the elements less than x (see below). The partition element x can appear anywhere in the “right partition”; it does not need to appear between the left and right partitions.

Similar problem : Partitioning a linked list around a given value and keeping the original order

Examples:

Input : 3 -> 5 -> 10 -> 2 -> 8 -> 2 -> 1 x = 5 Output : 1-> 2-> 2-> 3-> 5-> 10-> 8

If we don’t care about making the elements of the list “stable” then we can instead rearrange the elements by growing the list at the head and tail.

In this approach, we start a “new” list (using the existing nodes). Elements bigger than the pivot element are put at the tail and elements smaller are put at the head. Each time we insert an element, we update either the head or tail.

Below is C++ implementation of above idea.

// C++ program to partition a linked list around a // given value. #include<bits/stdc++.h> using namespace std; /* Link list Node */ struct Node { int data; struct Node* next; }; // A utility function to create a new node Node *newNode(int data) { struct Node* new_node = new Node; new_node->data = data; new_node->next = NULL; return new_node; } // Function to make a new list(using the existing // nodes) and return head of new list. struct Node *partition(struct Node *head, int x) { /* Let us initialize start and tail nodes of new list */ struct Node *tail = head; // Now iterate original list and connect nodes Node *curr = head; while (curr != NULL) { struct Node *next = curr->next; if (curr->data < x) { /* Insert node at head. */ curr->next = head; head = curr; } else // Append to the list of greater values { /* Insert node at tail. */ tail->next = curr; tail = curr; } curr = next; } tail->next = NULL; // The head has changed, so we need // to return it to the user. return head; } /* Function to print linked list */ void printList(struct Node *head) { struct Node *temp = head; while (temp != NULL) { printf("%d ", temp->data); temp = temp->next; } } // Driver program to run the case int main() { /* Start with the empty list */ struct Node* head = newNode(3); head->next = newNode(5); head->next->next = newNode(8); head->next->next->next = newNode(2); head->next->next->next->next = newNode(10); head->next->next->next->next->next = newNode(2); head->next->next->next->next->next->next = newNode(1); int x = 5; head = partition(head, x); printList(head); return 0; }

Output:

1 2 2 3 5 8 10

This article is contributed by **Mr. Somesh Awasthi**. 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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