Rotate List

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.

和求倒数第k个节点的那道题差不多，都是用fast和slow指针得到相应的位置，这道题需要注意一下k大于链表长度的情况。

class Solution {
public:
    ListNode *rotateRight(ListNode *head, int k) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(head == NULL || k == 0) return head;
        
           
        ListNode prehead(0);
        prehead.next = head;
        ListNode *p = head,*fast = head,*slow = &prehead;
        
        int len = 1;
        while(p->next != NULL)
        { 
            if(len++ == k) fast = p;
            p = p->next;
        }
        if(len == 1 || k%len == 0) return head;
        
        
        if(len <= k) {
            p = head;
            k %= len;
            int i = 1;
            while(p->next != NULL)
            { 
                if(i++ == k) fast = p;
                p = p->next;
            }
        }
        
        while(fast->next != NULL)
        { 
            fast = fast->next;
            slow = slow->next;
        }
             
        fast->next = head;
        head = slow->next;
        slow->next = NULL;
        return head;
    }
};

52 milli secs