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 1: Use function remove between m and n to do this.
public ListNode ReverseKGroup(ListNode head, int k)
{
var len = Utility.GetLinkedListLength(head);
var newhead = head;
for (var i = 0; i + k - 1 < len; i = i + k)
{
newhead = ReverseBetween(newhead, i, i + k - 1);
}
return newhead;
}
see reverse between m and n function
Solution 2: iterative solution. check if it has n nodes, if it has,reverse it one by one.
public ListNode ReverseKGroup(ListNode head, int k)
{
if (k == 1) return head;
var fakehead = new ListNode(-1) { next = head };
var start = fakehead;
var nodeToReverse = head;
while (true)
{
if (canmove(start, k) == null) break;
var count = 1;
while (count++ < k)
{
var tmp = start.next;
start.next = nodeToReverse.next;
nodeToReverse.next = nodeToReverse.next.next;
start.next.next = tmp;
}
start = nodeToReverse;
nodeToReverse = nodeToReverse.next;
}
return fakehead.next;
}
private ListNode canmove(ListNode head, int k)
{
var count = 0;
while (count++ < k)
{
if (head == null) return null;
head = head.next;
}
return head;
}
Solution 3: recursion solution.
private ListNode canmove(ListNode head, int k)
{
var count = 1;
while (count++ < k)
{
if (head == null) return null;
head = head.next;
}
return head;
}
public ListNode ReverseKGroup2(ListNode head, int k)
{
if (k == 1) return head;
var nextnode = canmove(head, k);
if (nextnode == null) return head;
var start = head.next;
head.next = ReverseKGroup(nextnode.next, k);
while (start.next != nextnode.next)
{
var tmp = start.next;
start.next = head;
head = start;
start = tmp;
}
start.next = head;
return start;
}