Description
Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.
k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.
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
Note:
Only constant extra memory is allowed.
You may not alter the values in the list’s nodes, only nodes itself may be changed.
Solution
// Iterative with constant memory
//O(n) O(1) 100% 24%
class Solution {
public ListNode reverseKGroup(ListNode head, int k) {
ListNode dummy = new ListNode(-1);
dummy.next = head;
ListNode curr = dummy;
ListNode end = traverseK(curr, k);
while (end != null) {
ListNode first = curr.next;
ListNode afterEnd = end.next;
end.next = null;
this.reverseList(first);
curr.next = end;
first.next = afterEnd;
curr = first;
end = traverseK(curr, k);
}
return dummy.next;
}
private ListNode reverseList(ListNode head) {
ListNode prev = null;
ListNode curr = head;
while (curr != null) {
ListNode next = curr.next;
curr.next = prev;
prev = curr;
curr = next;
}
return prev;
}
private ListNode traverseK(ListNode node, int k) {
int count = 0;
while (node.next != null) {
node = node.next;
count++;
if (count == k) break;
}
if (count == k) return node;
else return null;
}
}