LeetCode-61-Rotate List

算法描述：

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

解题思路：单链表的题目，画出图就可以解决了。注意边界条件，以及k值大于链表长度等问题。

    ListNode* rotateRight(ListNode* head, int k) {
        if(head==nullptr || k ==0) return head;
        ListNode* dup = new ListNode(-1);
        dup->next = head;
        ListNode* fast = dup;
        int count =0;
        while(fast->next!=nullptr){
            fast=fast->next;
            count++;
        }
        int n = count - k%count;
        ListNode* slow = dup;
        while(n>0){
            slow = slow->next;
            n--;
        }
        fast->next = dup->next;
        dup->next = slow->next;
        slow->next = nullptr;
        return dup->next;        
    }