Rotate List

题目：

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

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

分析：

这道题目有一个明显的陷阱，因为是right shift，所以k有可能大于数组的长度。于是乎遍历一次取得len，然后用k % len作为移动的位置。这里面有个特点，如果k % len == 0则可以return，因为相当于没有移动过。

解法：

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 == NULL || head->next == NULL) {
            return head;
        }

        if (k == 0) {
            return head;
        }

        // find length, in case k > len
        int len = 0;
        ListNode* curr = head;
        while (curr != NULL) {
            len++;
            curr = curr->next;
        }

        k = k % len;
        if (k == 0) {
            return head;
        }

        // start to work on problem
        ListNode* slow = head;
        ListNode* fast = head;
        for (int i = 0; i < k; i++) {
            fast = fast->next;
        }

        while (fast->next != NULL) {
            slow = slow->next;
            fast = fast->next;
        }

        ListNode* newhead = slow->next;
        slow->next = NULL;
        fast->next = head;

        return newhead;
    }
};