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.

这道题有点循环移位的意思

将链表向右移动k个节点，右边的节点放到左边。返回新的链表头结点

思路

遍历链表，找出长度length,对k取余，k = k % length,利用双指针一个frontPtr在前面k个节点，后一个behindPtr从head开始，frontPtr到链表结尾。forntPtr指向头结点，头结点改为behindPtr所指下一个节点。behindPtr下一个节点置为空。返回头结点即可

public class Solution {
    public ListNode rotateRight(ListNode head, int n) {
        if(head == null || head.next == null || n == 0)
            return head;
        ListNode behindPtr = head;
        ListNode frontPtr = head;
        
        int count = 0;
        ListNode temp = head;
        int length = 1;
        while(temp.next != null){
            length ++;
            temp = temp.next;
        }//while
        n = n % length;
        while(count < n && frontPtr.next != null){
            frontPtr = frontPtr.next;
            count ++;
        }//while
        while(frontPtr.next != null){
            frontPtr = frontPtr.next;
            behindPtr = behindPtr.next;
        }//while
        frontPtr.next = head;
        head = behindPtr.next;
        behindPtr.next = null;
                
        return head;
    }
}