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.

注意rotate回头的情况：

{1,2,3}, 5wrong:{1,2,3}right:{2,3,1}

突然想到，在白板上写程序，因为space limited，最好不要用无谓的{,}，而且尽量在同一行使得代码在保证可读性的情况下尽量短！！！

所以第二遍过的时候要注意怎么样代码更加的简洁

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *rotateRight(ListNode *head, int k) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (!k) return head;
        ListNode dummynode(0), *h = &dummynode,*p = head, *e = p;
        h->next = head;
        //rotate回来的情况，先得到总长度
        int length = 0;
        for(e = p;e;e = e->next){
            length ++;
        }
        //如果是满足mod的关系，可以直接返回,%的分母不能为0
        if (length && k%length == 0){
            return head;
        }
        int n = 0;
        for(e = p;e;e = e->next){
            if (n++ == k%length) break;
        }
        //因为k%length < length,所以e肯定为空,不过为了保险，下边判断的时候可以加入 while(p&e&e->next)
        while(p && e && e->next){
            p = p->next;
            e = e->next;
        }
        if (p) {
            h->next = p->next;
            p->next = NULL;
        }
        if (e) e->next = head;
        
        return h->next;
    }
};