LeetCode61. Rotate List（C++）

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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：旋转k步，即将后k%num节点放到表头，其中num为链表的节点个数。所以先遍历一遍得到链表的节点数，然后用双指针法，得到倒数k%num的节点的pre节点

class Solution {
public:
    ListNode* rotateRight(ListNode* head, int k) {
        int num=0;
        ListNode *p=head,*q;
        while(p!=NULL){
            p=p->next;
            num++;
        }
        if(num==0)
            return head;
        k%=num;
        p=head;q=head;
        for(int i=0;i<k;i++)
            p=p->next;
        while(p->next!=NULL){
            p=p->next;
            q=q->next;
        }
        p->next=head;
        head=q->next;
        q->next=NULL;
        return head;
    }
};