LeetCode -- Rotate List

题目描述：

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.

就是将倒数K个节点放在链表前。

思路：
1. 遍历链表判断长度len是否小于k，如果是，则k = k % len
2. 如果len = k或len < 2或k < 1，直接返回
3. 遍历链表，移动len - k - 1个位置，存当前位置为q
4. 将q之后的节点存入stack
5. 遍历stack，依次将节点设为头结点
6. 最后，将q.next指向Null

实现代码：

/**
* Definition for singly-linked list.
* public class ListNode {
*     public int val;
*     public ListNode next;
*     public ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode RotateRight(ListNode head, int k)
{
if(head == null){
return null;
}
var len = LenOf(head);
if(len < 2 || k == len || k < 1){
return head;
}
if(k > len){
k = k % len;
}

var stack = new Stack<int>();
var p = head;

for(var i = 0;i < len - k - 1;i ++){
p = p.next;
}

var q = p;

p = p.next;
while(p != null){
stack.Push(p.val);
p = p.next;
}
q.next = null;

while(stack.Count > 0)
{
var n = new ListNode(stack.Pop());
n.next = head;
head = n;
}

return head;
}

private int LenOf(ListNode head)
{
var p = head;
var len = 0;
while(p != null){
len ++;
p = p.next;
}
return len;
}
}