Given a linked list and a value x, partition it such that all nodes less than x come first, then all nodes with a value equal to x, and finally nodes with a value greater than or equal to x. The original relative order of the nodes in each of the three partitions should be preserved. The partition must work in place.
Examples:
Input: 1->4->3->2->5->2->3,
x = 3
Output: 1->2->2->3->3->4->5
Input: 1->4->2->10
x = 3
Output: 1->2->4->10
Input: 10->4->20->10->3
x = 3
Output: 3->10->4->20->10
To solve this problem we can use partition method of Quick Sort but this would not preserve the original relative order of the nodes in each of the two partitions.
Below is the algorithm to solve this problem :
- Initialize first and last nodes of below three linked lists as NULL.
- Linked list of values smaller than x.
- Linked list of values equal to x.
- Linked list of values greater than x.
- Now iterate through the original linked list. If a node’s value is less than x then append it at the end of the smaller list. If the value is equal to x, then at the end of the equal list. And if a value is greater, then at the end of the greater list.
- Now concatenate three lists.
Below is the implementation of the above idea.
Javascript
<script>// Javascript program to partition a // linked list around a given value. // Link list Node class Node { constructor() { this.data = 0; this.next = null; }} // A utility function to create // a new nodefunction newNode(data) { var new_node = new Node(); new_node.data = data; new_node.next = null; return new_node;}// Function to make two separate lists // and return head after concatenatingfunction partition(head, x) { /* Let us initialize first and last nodes of three linked lists 1) Linked list of values smaller than x. 2) Linked list of values equal to x. 3) Linked list of values greater than x. */ var smallerHead = null, smallerLast = null; var greaterLast = null, greaterHead = null; var equalHead = null, equalLast = null; // Now iterate original list and // connect nodes of appropriate // linked lists. while (head != null) { // If current node is equal to x, // append it to the list of x values if (head.data == x) { if (equalHead == null) equalHead = equalLast = head; else { equalLast.next = head; equalLast = equalLast.next; } } // If current node is less than X, // append it to the list of smaller // values else if (head.data < x) { if (smallerHead == null) smallerLast = smallerHead = head; else { smallerLast.next = head; smallerLast = head; } } // Append to the list of greater values else { if (greaterHead == null) greaterLast = greaterHead = head; else { greaterLast.next = head; greaterLast = head; } } head = head.next; } // Fix end of greater linked list to NULL // if this list has some nodes if (greaterLast != null) greaterLast.next = null; // Connect three lists // If smaller list is empty if (smallerHead == null) { if (equalHead == null) return greaterHead; equalLast.next = greaterHead; return equalHead; } // If smaller list is not empty // and equal list is empty if (equalHead == null) { smallerLast.next = greaterHead; return smallerHead; } // If both smaller and equal list // are non-empty smallerLast.next = equalHead; equalLast.next = greaterHead; return smallerHead;}// Function to print linked list function printList(head) { var temp = head; while (temp != null) { document.write(temp.data + " "); temp = temp.next; }}// Driver code// Start with the empty list var head = newNode(10);head.next = newNode(4);head.next.next = newNode(5);head.next.next.next = newNode(30);head.next.next.next.next = newNode(2);head.next.next.next.next.next = newNode(50);var x = 3;head = partition(head, x);printList(head);// This code is contributed by aashish1995 </script> |
Output:
2 10 4 5 30 50
Time Complexity: O(n) where n is the size of the given linked list.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
Please refer complete article on Partitioning a linked list around a given value and keeping the original order for more details!
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