Given a linked list and two positions ‘m’ and ‘n’. The task is to rotate the sublist from position m to n, to the right by k places. Examples:
Input: list = 1->2->3->4->5->6, m = 2, n = 5, k = 2 Output: 1->4->5->2->3->6 Rotate the sublist 2 3 4 5 towards right 2 times then the modified list are: 1 4 5 2 3 6 Input: list = 20->45->32->34->22->28, m = 3, n = 6, k = 3 Output: 20->45->34->22->28->32 Rotate the sublist 32 34 22 28 towards right 3 times then the modified list are: 20 45 34 22 28 32
Approach: For rotating the given sublist that extends from m to n element, move the list from (n-k+1)th to nth node to starting of sub-list to finish the rotation. If k is greater than size of sublist then we will take its modulo with size of sublist. So traverse through list using a pointer and a counter and we will save (m-1)th node and later make it point to (n-k+1)th node and hence bring (n-k+1)th node to the start(front) of sublist. Similarly we will save mth node and later make nth node point to it. And for keeping rest of list intact we will make (n-k)th node point to next node of n (maybe NULL). And finally we will get the k times right rotated sublist. Below is the implementation of the above approach:
Java
// Java implementation of the above approachimport java.util.*;class Solution{ // Definition of node of linkedliststatic class ListNode { int data; ListNode next;} // This function take head pointer of list, start and// end points of sublist that is to be rotated and the// number k and rotate the sublist to right by k places.static void rotateSubList(ListNode A, int m, int n, int k){ int size = n - m + 1; // If k is greater than size of sublist then // we will take its modulo with size of sublist if (k > size) { k = k % size; } // If k is zero or k is equal to size or k is // a multiple of size of sublist then list // remains intact if (k == 0 || k == size) { ListNode head = A; while (head != null) { System.out.print( head.data); head = head.next; } return; } ListNode link = null; // m-th node if (m == 1) { link = A; } // This loop will traverse all node till // end node of sublist. ListNode c = A; // Current traversed node int count = 0; // Count of traversed nodes ListNode end = null; ListNode pre = null; // Previous of m-th node while (c != null) { count++; // We will save (m-1)th node and later // make it point to (n-k+1)th node if (count == m - 1) { pre = c; link = c.next; } if (count == n - k) { if (m == 1) { end = c; A = c.next; } else { end = c; // That is how we bring (n-k+1)th // node to front of sublist. pre.next = c.next; } } // This keeps rest part of list intact. if (count == n) { ListNode d = c.next; c.next = link; end.next = d; ListNode head = A; while (head != null) { System.out.print( head.data+" "); head = head.next; } return; } c = c.next; }} // Function for creating and linking new nodesstatic ListNode push( ListNode head, int val){ ListNode new_node = new ListNode(); new_node.data = val; new_node.next = (head); (head) = new_node; return head;} // Driver codepublic static void main(String args[]){ ListNode head = null; head =push(head, 70); head =push(head, 60); head =push(head, 50); head =push(head, 40); head =push(head, 30); head =push(head, 20); head =push(head, 10); ListNode tmp = head; System.out.print("Given List: "); while (tmp != null) { System.out.print( tmp.data + " "); tmp = tmp.next; } System.out.println(); int m = 3, n = 6, k = 2; System.out.print( "After rotation of sublist: "); rotateSubList(head, m, n, k); }}// This code is contributed// by Arnab Kundu |
Given List: 10 20 30 40 50 60 70 After rotation of sublist: 10 20 50 60 30 40 70
Time Complexity: O(n) where n is no of node in given linked list
Auxiliary Space: O(1) since it is using constant space
Please refer complete article on Rotate the sub-list of a linked list from position M to N to the right by K places for more details!
