题目：
Given a linked list, rotate the list to the right by k places, where k is non-negative.
Example 1:
Input: 1->2->3->4->5->NULL, k = 2 Output: 4->5->1->2->3->NULL Explanation: rotate 1 steps to the right: 5->1->2->3->4->NULL rotate 2 steps to the right: 4->5->1->2->3->NULL
Example 2:
Input: 0->1->2->NULL, k = 4 Output:2->0->1->NULL
Explanation:
rotate 1 steps to the right: 2->0->1->NULL
rotate 2 steps to the right: 1->2->0->NULL
rotate 3 steps to the right:0->1->2->NULL
rotate 4 steps to the right:2->0->1->NULL
解题：
1. 遍历链表，计算链表个数len，并将链表头尾相连；
2. 从head遍历节点，找到要到第 len-(k%len) 个的前一个节点cur ;
3. cur->next 做为新的头节点返回。
实现：
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
struct ListNode* rotateRight(struct ListNode* head, int k){
int len = 0;
struct ListNode *cur=head;
if (NULL == head){
return head;
}
len = 1;
while (cur->next!=NULL){
cur=cur->next;
len += 1;
}
cur->next = head;
int r=len-(k%len);
for (int i=0; i<r; i++){
cur = cur->next;
}
struct ListNode *newhead;
newhead = cur->next;
cur->next = NULL;
return newhead;
}
Runtime: 0 ms
Memory Usage: 7.3 MB
时间复杂度O(n)， 空间复杂度O(1)
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