标签:链表旋转 linkedlist 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.
本文的题意就是循环将后面的n个结点移动到前面去,所以,n有可能是大于链表的长度的,这是一个小小的陷阱。然后就是很简单的细节了,有点脑残,提交了好多次。
public ListNode rotateRight(ListNode head, int n) {
if(n<=0||head==null)
return head;
int len=1,index=0;
ListNode pListNode=head,qListNode=null,tailListNode=null;
while(pListNode.next!=null)
{
len++;
pListNode=pListNode.next;
}
tailListNode=pListNode;
n=n%len;
if(n==0)
return head;
index=len-n;
index--;
pListNode=head;
for(int i=0;i<index;i++)
{
pListNode=pListNode.next;
}
qListNode=pListNode.next;
pListNode.next=null;
pListNode=head;
head=qListNode;
tailListNode.next=pListNode;
return head;
}标签:链表旋转 linkedlist rotate list
原文地址:http://blog.csdn.net/mnmlist/article/details/43371871