Description
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
Two-pointers
由于k可能超出链表长度，所以先得到链表长度然后k对len取模，再进行快慢指针计算。
/**
* 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 len = getLength(head);
k %= len;
if (k <= 0) return head;
ListNode slow = head;
ListNode fast = head;
while (k-- > 0) {
fast = fast.next;
}
while (fast != null && fast.next != null) {
slow = slow.next;
fast = fast.next;
}
ListNode newHead = slow.next;
fast.next = head;
slow.next = null;
return newHead;
}
public int getLength(ListNode head) {
int len = 0;
while (head != null) {
head = head.next;
++len;
}
return len;
}
}
