[leetcode 25] Reverse Nodes in k-Group

Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.

If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.

You may not alter the values in the nodes, only nodes itself may be changed.

Only constant memory is allowed.

For example,
Given this linked list:

1->2->3->4->5

For k = 2, you should return:

2->1->4->3->5

For k = 3, you should return:

3->2->1->4->5

[Solution]

每次访问连续的k个节点，将k个节点逆置。

注意首位节点的衔接。

ListNode *reverseKGroup(ListNode *head, int k)
{
if (k <= 1 || head == NULL)
return head;
ListNode *dummy = new ListNode(INT_MIN);
dummy->next = head;
ListNode *p, *q, *tmp_head, *tmp_tail;
int i = 0;

p = dummy;
tmp_head = dummy;
while (p != NULL)
{
if (i == k)
{
q = p->next;
p->next = NULL;
tmp_tail = tmp_head->next;
tmp_head->next = reverse(tmp_head->next);
p = tmp_tail;
p->next = q;
tmp_head = p;
i = 0;
}
else
{
i++;
p = p->next;
}
}

head = dummy->next;
delete dummy;
return head;
}

ListNode *reverse(ListNode *head)
{
ListNode *p = NULL, *q, *r;

q = head;
if (q == NULL)
return NULL;
r = q->next;
while (r != NULL)
{
q->next = p;
p = q;
q = r;
r = r->next;
}
q->next = p;

return q;
}