给定一个链表，旋转链表，将链表每个节点向右移动 k 个位置，其中 k 是非负数。
示例 1:
代码语言：javascript复制输入: 1->2->3->4->5->NULL, k = 2
输出: 4->5->1->2->3->NULL
解释:
向右旋转 1 步: 5->1->2->3->4->NULL
向右旋转 2 步: 4->5->1->2->3->NULL示例 2:
代码语言：javascript复制输入: 0->1->2->NULL, k = 4
输出: 2->0->1->NULL
解释:
向右旋转 1 步: 2->0->1->NULL
向右旋转 2 步: 1->2->0->NULL
向右旋转 3 步: 0->1->2->NULL
向右旋转 4 步: 2->0->1->NULL解：双指针
代码语言：javascript复制/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode rotateRight(ListNode head, int k) {
if (head == null || head.next == null || k <= 0) {
return head;
}
int length = len(head);
int realK = k % length;
//移动0个位置
if (0 == realK) {
return head;
}
ListNode fast = head;
ListNode slow = head;
for (int i = 0; i < realK; i ) {
fast = fast.next;
}
while (fast.next != null) {
fast = fast.next;
slow = slow.next;
}
ListNode tmp = slow;
slow = slow.next;
tmp.next = null;
fast.next = head;
return slow;
}
private int len(ListNode head) {
int count = 0;
while (head != null) {
count ;
head = head.next;
}
return count;
}
}
