给定一个链表，旋转链表，将链表每个节点向右移动 k 个位置，其中 k 是非负数。
示例 1:
输入: 1->2->3->4->5->NULL, k = 2 输出: 4->5->1->2->3->NULL 解释: 向右旋转 1 步: 5->1->2->3->4->NULL 向右旋转 2 步: 4->5->1->2->3->NULL
示例 2:
输入: 0->1->2->NULL, k = 4 输出:2->0->1->NULL解释: 向右旋转 1 步: 2->0->1->NULL 向右旋转 2 步: 1->2->0->NULL 向右旋转 3 步:0->1->2->NULL向右旋转 4 步:2->0->1->NULL
/**
* 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) {
if(!head) return NULL;
int n = 0;
ListNode *cur = head;
while(cur){
++n;
cur = cur->next;
}
k%=n;
ListNode *fast = head, *slow = head;
for(int i = 0; i < k; i++){
if(fast) fast = fast->next;
}
if(!fast) return head;
while(fast->next){
fast = fast->next;
slow = slow->next;
}
fast->next = head;
fast = slow->next;
slow->next = NULL;
return fast;
}
};/**
* 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) {
if(!head) return NULL;
int n = 1;
ListNode *cur = head;
while(cur->next){
++n;
cur = cur->next;
}
cur->next = head;
int m = n-k%n;
for(int i = 0; i < m; i++){
cur = cur->next;
}
ListNode *newhead = cur->next;
cur->next = NULL;
return newhead;
}
};