61. Rotate List

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */

static int wing=[]()
{
    std::ios::sync_with_stdio(false);
    cin.tie(NULL);
    return 0;
}();

class Solution 
{
public:
    ListNode* rotateRight(ListNode* head, int k) 
    {
        int len=0;
        ListNode *p=head;
        while(p!=NULL)
        {
            p=p->next;
            len++;
        }
        if(len==0)
            return head;
        k%=len;
        if(k==0)
            return head;
        ListNode *fast=head,*slow=head;
        while(k>0)
        {
            fast=fast->next;
            k--;
        }
        while(fast->next!=NULL)
        {
            slow=slow->next;
            fast=fast->next;
        }
        fast->next=head;
        fast=slow->next;
        slow->next=NULL;
        return fast;
    }
};

循环移动链表元素，关键还是找倒数第k个节点，找到之后，保存下一个节点作为返回值，原末尾节点连在头结点，倒数第k个节点next域置空作为新尾节点。完事儿