[Leetcode] Rotate List

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.

Solution 1:

首先从head开始跑，直到最后一个节点，这时可以得出链表长度len。然后将尾指针指向头指针，将整个圈连起来。

这个既然要rotate，不如先连起来loop，这样怎么也不怕null pointer exception了。然后再找到该断开的地方断开。

/**
 * 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) {
        if(head==null||head.next==null)
            return head;
        ListNode temp=head;
        int count=1;
        while(temp.next!=null){
            count++;
            temp=temp.next;
        }
        temp.next=head;
        n%=count;
        ListNode p=head;
        for(int i=0;i<count-n-1;i++){
            p=p.next;
        }
        temp=p.next;
        p.next=null;
        return temp;
    }
}

Solution 2:

利用快慢指针来做。

/**
 * 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) {
        if(head==null||head.next==null)
        return head;
        ListNode dummy=new ListNode(-1);
        dummy.next=head;
        ListNode fast=dummy;
        ListNode slow=dummy;
        ListNode temp=head;
        int count=1;
        while(temp.next!=null){
            count++;
            temp=temp.next;
        }
        n%=count;
        while(fast.next!=null&&n!=0){
            fast=fast.next;
            n--;
        }
        if(n!=0||fast.next==null)
            return dummy.next;
        while(fast.next!=null){
            fast=fast.next;
            slow=slow.next;
        }
        fast.next=dummy.next;
        dummy.next=slow.next;
        slow.next=null;
        
        return dummy.next;
    }
}