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.
Follow up:
Could you solve the problem in O(1) extra memory space? You may not alter the values in the list's nodes, only nodes itself may be changed.
Example 1:
Input: head = [1,2,3,4,5], k = 2
Output: [2,1,4,3,5]
Example 2:
Input: head = [1,2,3,4,5], k = 3
Output: [3,2,1,4,5]
Example 3:
Input: head = [1,2,3,4,5], k = 1
Output: [1,2,3,4,5]
Example 4:
Input: head = [1], k = 1
Output: [1]
Constraints:
- The number of nodes in the list is in the range sz.
- 1 <= sz <= 5000
- 0 <= Node.val <= 1000
- 1 <= k <= sz
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode reverseKGroup(ListNode head, int k) {
if(k==1||head==null){
return head;
}
ListNode ff_prev = null,ff=head,ss_prev=head,ss=null;
while(true){
int z = ss==null ? k-1 :k;
while(ss_prev!=null && z-->0){
ss_prev = ss_prev.next;
}
if(ss_prev==null)
break;
ss=ss_prev.next;
ss_prev.next = null;
ListNode[] arr = reverse(ff);
if(ff_prev==null){
head=arr[0];
}else{
ff_prev.next = arr[0];
}
ff.next = ss;
ss_prev = arr[1];
ff_prev = ss_prev;
ff=ss;
}
return head;
}
private ListNode[] reverse(ListNode head){
ListNode pre = head;
ListNode now = head;
while(now!=null){
ListNode next = now.next;
now.next = pre;
pre = now;
now = next;
}
return new ListNode[]{pre, head};
}
}class Solution {
public ListNode reverseKGroup(ListNode head, int k) {
//checking size
ListNode sizeNode = head;
int size = 0;
while(sizeNode!=null){
sizeNode= sizeNode.next;
size++;
}
if(head==null || head.next == null || size<k )
return head;
ListNode curr = head;
ListNode prev= null;
ListNode nextNode ;
int count = 0;
while(curr!=null && count < k) {
nextNode= curr.next;
curr.next = prev;
prev = curr;
curr = nextNode;
count++;
}
ListNode temp = prev;
while(temp.next!=null){
temp = temp.next;
}
temp.next = reverseKGroup(curr,k);
return prev;
}
}