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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.
思路：从length－k(k%length)处切段，把前半段接到后半段。1.扫一遍，扫到末尾的时候把末尾与头连上。2.扫到length－k(k%length)处，也就是3，3.next变成head，3.next变成null
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ public class Solution { public ListNode rotateRight(ListNode head, int k) { if(head==null||k==0) { return head; } ListNode node=head; int size=1; while(node.next!=null) { node=node.next; size++; } node.next=head; k%=size; for(int i=0;i<size-k;i++) { node=node.next; } head=node.next; node.next=null; return head; } }
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原文地址：http://www.cnblogs.com/Machelsky/p/5940754.html
