题目
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
.
先得到链表长度len,n对len取余就是需要右转长度(也可以不要长度,就是指针移动n次找到剪切位置,但是如果n过大就悲剧了),然后找到剪切位置,这里依然用个哨兵简化代码。
代码
public class RotateList { public ListNode rotateRight(ListNode head, int n) { if (head == null || n == 0) { return head; } ListNode tail = head; int len = 1; while (tail.next != null) { tail = tail.next; ++len; } n = len - n % len; if (n == len) { return head; } ListNode dummy = new ListNode(0); dummy.next = head; head = dummy; for (int i = 0; i < n; ++i) { head = head.next; } tail.next = dummy.next; dummy.next = head.next; head.next = null; return dummy.next; } }
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原文:http://blog.csdn.net/perfect8886/article/details/21414629