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This approach involves converting the linked list into a circular list and then breaking it at the correct point to achieve the required rotation.
By making the linked list a circular one, you can easily traverse k nodes from the end to set the new head and make necessary detachments.
Time Complexity: O(n) because we pass through the list twice, once to find the length and once to find the new head.
Space Complexity: O(1) since we only use a constant amount of extra space.
1class ListNode {
2 int val;
3 ListNode next;
4 ListNode(int x) { val = x; }
5}
6
7class Solution {
8 public ListNode rotateRight(ListNode head, int k) {
9 if (head == null || k == 0) return head;
10 ListNode curr = head;
11 int length = 1;
12 while (curr.next != null) {
13 curr = curr.next;
14 length++;
15 }
16 curr.next = head; // Make it circular
17 k = k % length;
18 for (int i = 0; i < length - k; i++) {
19 curr = curr.next;
20 }
21 head = curr.next;
22 curr.next = null;
23 return head;
24 }
25
26 public static void main(String[] args) {
27 ListNode head = new ListNode(1);
28 head.next = new ListNode(2);
29 head.next.next = new ListNode(3);
30 head.next.next.next = new ListNode(4);
31 head.next.next.next.next = new ListNode(5);
32
33 Solution sol = new Solution();
34 head = sol.rotateRight(head, 2);
35
36 while (head != null) {
37 System.out.print(head.val + " ");
38 head = head.next;
39 }
40 }
41}This Java solution repeats the same steps as the C implementation. Make the list circular by connecting the last node back to the head. Calculate k % length to find the effective number of moves. Finally, walk through the circular list to form the desired result, closing it correctly upon completion.
In this approach, we utilize a two-pointer technique. We advance one pointer ahead by k nodes and then move another pointer from the start concurrently until the first pointer reaches the list's end. This way, the second pointer will eventually land on the node that will become the new tail.
Time Complexity: O(n) because it requires traversal for counting.
Space Complexity: O(1) as no extra space is used.
1public class ListNode {
2 public int val;
3 public ListNode next;
4 public ListNode(int val = 0, ListNode next = null) {
this.val = val;
this.next = next;
}
}
public class Solution {
public ListNode RotateRight(ListNode head, int k) {
if (head == null || k == 0) return head;
ListNode fast = head, slow = head;
int length = 1;
while (fast.next != null) {
fast = fast.next;
length++;
}
fast.next = head; // form circular
k %= length;
for (int i = 0; i < length - k; i++) {
slow = slow.next;
fast = fast.next;
}
head = slow.next;
fast.next = null;
return head;
}
public static void Main() {
ListNode head = new ListNode(1, new ListNode(2, new ListNode(3, new ListNode(4, new ListNode(5)))));
Solution sol = new Solution();
ListNode newHead = sol.RotateRight(head, 2);
ListNode curr = newHead;
while (curr != null) {
Console.Write(curr.val + " ");
curr = curr.next;
}
}
}
In this C# solution, the logic is similar to the previous languages. After determining the length and connecting the end, the fast and slow pointers identify the correct division point. This method makes efficient use of mid-list pointer adjustments to achieve rotation.