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.
一个小错误, 简直可以说是笔误, 浪费我近两个小时
悲夫~
启示:
当判断是否为NULL,时
如果是指针, 写成NULL == p 形式
如果是整形, 写成0 == n;
while(!np->next) 这里不是!, 去掉!
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
/*ListNode *kthNode(ListNode *head, int k)
{
if(!head||k==0||!head->next)
return head;
ListNode *np = head;
ListNode *temp;
int len = 1;
while(!np->next)
{
len++;
np = np->next;
}
np->next = head;
k = k%len;
k = len - k ;
while(k--)
np = np->next;
temp = np->next;
np->next = NULL;
return temp;
}
*/
ListNode *rotateRight(ListNode *head, int k) {
if(!head||k==0||!head->next)
return head;
ListNode *np = head;
ListNode *temp;
int len = 1;
while(np->next)
{
len++;
np = np->next;
}
np->next = head;
k = k%len;
k = len - k ;
while(k--)
np = np->next;
temp = np->next;
np->next = NULL;
return temp;
}
};