给定一个链表,旋转链表,将链表每个节点向右移动 k 个位置,其中 k 是非负数。
基本思路是每次循环先找到倒数第二节点的位置,然后把头结点的next指向最后一个节点,倒数第二节点的next指向NULL
需要注意的是,实际上K=1和K=链表长度+1时,返回的结果是一致的,需要对K作取余处理
ListNode* Solution::rotateRight(ListNode* head, int k)
{
ListNode *pnewhead = new ListNode(0);
ListNode *ptemp;
ListNode *plength = head;
unsigned int index = 0;
unsigned int length = 0;
unsigned int loop = 0;
if((head == NULL)||(head->next == NULL))
{
return head;
}
while(plength != NULL)
{
plength = plength->next;
length ++;
}
if(k%length == 0)
{
loop = length;
}
else
{
loop = k%length;
}
pnewhead->next = head;
for(index = 0; index < loop; index ++)
{
ListNode *plast = pnewhead->next;
while(plast->next->next != NULL)
{
plast = plast->next;
}
ptemp = pnewhead->next;
pnewhead->next = plast->next;
plast->next->next = ptemp;
plast->next = NULL;
}
return pnewhead->next;
}