61. 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.

解题思路为two pointers。

首先用一个fast指针走到表尾，并记下表长len。 然后用一个slow指针走(len-k%len)步，到达断点。 最后将fast与原表头连起来，slow的后一个node设为新表头，并把slow.next设为null。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode rotateRight(ListNode head, int k) {
        if(head == null || head.next == null) return head;
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode fast = dummy;
        int len = 0;
        while(fast.next != null) {
            fast = fast.next;
            len++;
        }
        ListNode slow = dummy;
        for(int i = len - k % len; i > 0; i--) {
            slow = slow.next;
        }
        fast.next = dummy.next;
        dummy.next = slow.next;
        slow.next = null;
        return dummy.next;
    }
}