<LeetCode OJ> 61. Rotate List

Total Accepted: 69038 Total
Submissions: 300765 Difficulty: Medium

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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(E) Rotate Array

分析：

具体分析见代码，比较简单！

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
//1,首先获取链表的长度，再根据k计算正序的新链表头
//2，建立连接
class Solution {
public:
ListNode* rotateRight(ListNode* head, int k) {
if(head==NULL || k==0)
return head;
int len=0;
ListNode* pmove=head;
ListNode* pend=NULL;
ListNode* pnewhead=NULL;
//1，获取链表长度
while(pmove)
{
len++;
pend=pmove;//记录尾节点
pmove=pmove->next;
}

//2，找到新头的位置
k=k%len;//取余
if(k==0)
return head;
int nlen=len-k;
pmove=head;
while(--nlen)
pmove=pmove->next;
//3，建立连接
pnewhead=pmove->next;
pmove->next=NULL;
pend->next=head;
return pnewhead;
}
};

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原文地址：http://blog.csdn.net/ebowtang/article/details/51422358

原作者博客：http://blog.csdn.net/ebowtang

本博客LeetCode题解索引：/article/3664871.html