Leetcode–Rotate List

The Problem:

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

For example:
Given 1->2->3->4->5->NULL and k = 2,
return 4->5->1->2->3->NULL.


  1. Two pointers with one k elements ahead, so when the faster reach null, the slower one is the new head.
  2. Attention that k could be bigger than the length, so within the iteration, if fast reach to null, re-initialize to head.
  3. Mistake I made:
    1. Cases no need to proceed : 1. 0 node, 2. 1 node, 3. n=0
    2. After the while loop, if fast reach the end, means n is the length of list, just return the original head.

Java Solution:

 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
public class Solution {
    public ListNode rotateRight(ListNode head, int n) {
        //--!!!---cases no need to proceed: 0 node; 1 node; n=0
        if(head==null || head.next == null || n==0) return head;
        ListNode fast = head;
            if(fast == null)//---n could be bigger than length
                fast = head;
            fast = fast.next;
        if(fast == null) // --!!! n is the length of list, just the same after rotation
            return head;
        ListNode slow = head;
        while(fast!= null && fast.next!=null){
            fast = fast.next;
            slow = slow.next;
        ListNode newHead = slow.next;
        slow.next = null;
        fast.next = head;
        return newHead;

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