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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.
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解答：
不是很懂这道题的通过率为何如此之低，其实只要考虑三种边界情况就可以了，一是空链表的时候，二是k是链表节点数整数倍的时候，三是k大于链表节点数的时候……
/** * Definition for singly-linked list. * struct ListNode { * int val; * struct ListNode *next; * }; */ struct ListNode* rotateRight(struct ListNode* head, int k) { int count = 0, i; struct ListNode *pNode = head, *tmp; if(NULL == head) return; while(NULL != pNode){ count++; pNode = pNode->next; } pNode = head; for(i = 0; i < count - k % count - 1; i++){ pNode = pNode->next; } tmp = pNode->next; pNode->next = NULL; if(NULL == tmp) return head; pNode = tmp; while(NULL != pNode->next){ pNode = pNode->next; } pNode->next = head; return tmp; }
标签：scribe where span str log 大于 ini this next
原文地址：http://www.cnblogs.com/YuNanlong/p/6055464.html
