Amazon

Bloomberg

2020-10-07

## 61. Rotate List

### Question:

Given a linked list, rotate the list to the right by k places, where k is non-negative.

#### Example 1:

``````Input: 1->2->3->4->5->NULL, k = 2
Output: 4->5->1->2->3->NULL
Explanation:
rotate 1 steps to the right: 5->1->2->3->4->NULL
rotate 2 steps to the right: 4->5->1->2->3->NULL
``````

#### Example 2:

``````Input: 0->1->2->NULL, k = 4
Output: 2->0->1->NULL
Explanation:
rotate 1 steps to the right: 2->0->1->NULL
rotate 2 steps to the right: 1->2->0->NULL
rotate 3 steps to the right: 0->1->2->NULL
rotate 4 steps to the right: 2->0->1->NULL
``````

### Solution:

There are two different ways of using the pointers
Time Complexity: O(n) Both are using the same idea: finding the connection point and relink the list.

``````/**
* Definition for singly-linked list.
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode() {}
*     ListNode(int val) { this.val = val; }
*     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode rotateRight(ListNode head, int k) {
if (head == null || head.next == null) return head;
int counter = 1;
ListNode tail = head;
while (tail.next != null) {
tail = tail.next;
counter++;
}
k = k % counter;
if (k == 0) {
}
ListNode newTail = head;
for (int i = 0; i < counter - k - 1; i++){
newTail = newTail.next;
}
ListNode result = newTail.next;
newTail.next = null;

return result;
}
}
``````

Use two pointers, one is faster than the other with the gap of k.

``````/**
* Definition for singly-linked list.
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode() {}
*     ListNode(int val) { this.val = val; }
*     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode rotateRight(ListNode head, int k) {
if (head == null || head.next == null) return head;
int counter = 0;

// Two Pointers
ListNode fast = head;
ListNode slow = head;
while (fast != null) {
fast = fast.next;
counter++;
}
k = k % counter;

// The difference between two pointers is k
for (int i = 0; i < k; i++) {
fast = fast.next;
}

// Two pointer move to the end
while (fast.next != null) {
fast = fast.next;
slow = slow.next;
}

// slow is pointing to the tail and fast is pointing to the connection point
// slow.next is the head