Rotate List

Given a list, rotate the list to the right by k places, where k is non-negative.

Have you met this question in a real interview? 

Yes

Example

Given 1->2->3->4->5 and k = 2, return 4->5->1->2->3.

贴一个错的答案，没有考虑超过长度的问题。。。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    /**
     * @param head: the list
     * @param k: rotate to the right k places
     * @return: the list after rotation
     */
    ListNode *rotateRight(ListNode *head, int k) {
        // write your code here
        if (!head || !head->next)
            return head;
        ListNode dummy(0);
        dummy.next=head;
        head=&dummy;
        for(int i=0;i<k;i++){
            if (!head)
                return 0;  
            head=head->next;
        }
        ListNode* slow=&dummy;
        while(head->next){
            slow=slow->next;
            head=head->next;
        }
        ListNode* tmp=slow->next;
        slow->next=0;
        tmp->next=dummy.next;
        return tmp;
    }
};

这个解得测试结果：

Input

17->75->80->87->44->45->75->86->74->20->null, 19

Output

null

Expected

75->80->87->44->45->75->86->74->20->17->null

所以需要一个余数，先算个n，不过写了半天还是有bug，一怒之下换个思路，用环，先把首位相连，然后再重新断开。。。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    /**
     * @param head: the list
     * @param k: rotate to the right k places
     * @return: the list after rotation
     */
    ListNode *rotateRight(ListNode *head, int k) {
        // write your code here
        if(!head || !head->next)
            return head;
        int count=1;
        ListNode* runNode=head;
        while(runNode->next){
            count++;
            runNode=runNode->next;
        }
        runNode->next=head;
        runNode=head;
        for (int i=1; i<count-k%count; i++){
            runNode=runNode->next;
        }
        head=runNode->next;
        runNode->next=0;
        return head;
    }
};