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.
巨没劲的一道题,当k>length 时,我以为origin list 不动就行,结果好些case过不了,看了别人的代码才知道要 k%= length 真心想不通为什么,也懒得弄这种恶心的东西。面试遇到这种题目直接问面试官这种奇葩情况怎么处理就行。
/** * 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(head == NULL || k == 0) return head; ListNode *p, *q; p = head;q= p; while(q &&--k){ q = q->next; } if(k >0 || NULL == q) return head; ListNode *pre = NULL; while(q->next){ pre = p; p = p->next; q = q->next; } q ->next = head; head = p; pre ->next = NULL; return head; } };