[LeetCode]Rotate List

链表题again. 实现链表节点的循环移位. 注意k的边界范围。

思路：

1. 得到链表长度N，指针end指向尾节点

2. 找到新链表头节点在老链表中的前驱节点，记为p. p = N-k%N-1

3. end指向原链表head节点，此时链表为环状。更新链表头节点位置，head指向p->next。将p->next置为NULL，成为尾节点。

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.

代码

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *rotateRight(ListNode *head, int k) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        if(head == NULL || k==0)
            return head;
        
        ListNode *end = head;
        int N = 1;
        while(end->next)
        {
            ++N;
            end = end->next;
        }
        int p = N-k%N-1;
        ListNode *pre = head;
        if(p<0)
            return head;
        else
        {
            for(int i=0; i<p; ++i)
                pre = pre->next;
            end->next = head;
            head = pre->next;
            pre->next = NULL;
        }
        return head;
    }
};