[leetcode-61-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.

思路：

首先求出list的长度length，然后将最后一个结点指向初始的头结点，

然后从该初始结点出发，指针向后移动length-k次，

找到新链表的尾结点，然后把它的next置为NULL。

ListNode* rotateRight(ListNode* head, int k){ListNode* result;if (k == 0 || head == NULL) return head;int length = 1;ListNode* temp = head;while (head != NULL && head->next != NULL){head = head->next;length++;}k %= length;head->next = temp;//head 目前为最后一个结点的指针 指向初始时候的头部结点for (int i = 0; i < length - k-1;i++){temp = temp->next;}result = temp->next;temp->next = NULL;return result;}

转载于:https://www.cnblogs.com/hellowooorld/p/6674159.html

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