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.

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

**Java Solution**

public ListNode reverseKGroup(ListNode head, int k) { if(head==null||k==1) return head; ListNode fake = new ListNode(0); fake.next = head; ListNode prev = fake; int i=0; ListNode p = head; while(p!=null){ i++; if(i%k==0){ prev = reverse(prev, p.next); p = prev.next; }else{ p = p.next; } } return fake.next; } |

public ListNode reverse(ListNode prev, ListNode next){ ListNode last = prev.next; ListNode curr = last.next; while(curr != next){ last.next = curr.next; curr.next = prev.next; prev.next = curr; curr = last.next; } return last; } |

We can write the reverse method differently like the following. I personally it is more understandable.

private ListNode reverse(ListNode prev, ListNode next){ ListNode p1 = prev.next; ListNode p2 = p1.next; while(p2 != next){ ListNode t = p2.next; p2.next = p1; p1 = p2; p2 = t; } ListNode rNode = prev.next; prev.next.next = next; prev.next = p1; return rNode; } |

Hi Guys

I created a small video explaining the recursion approach, you can check this out

https://www.youtube.com/watch?v=u1Ny3aPknv0

Recursive Java Solution:https://www.youtube.com/watch?v=8zuSdN4ru7M

A JavaScript Solution:

Chinese: https://www.youtube.com/watch?v=Mt2ID8xuR5Q

Facebook: https://www.facebook.com/groups/2094071194216385/

Reverse K group using stack.

public void reverseKGroup(int k){

Stack tempStack = new Stack();

Node current = root;

Node previous = null;

int i = 0;

while(current != null){

if(i!=0 && i%k == 0){

Node curr = tempStack.pop();

if(i==k){

root = curr;

}else{

previous.next = curr;

}

while(!tempStack.empty()){

Node n = tempStack.pop();

curr.next = n;

curr = n;

}

curr.next = current;

previous = curr;

}

tempStack.push(current);

current = current.next;

i++;

}

if(i%k == 0){

Node curr = tempStack.pop();

previous.next = curr;

while(!tempStack.empty()){

Node n = tempStack.pop();

curr.next = n;

curr = n;

}

curr.next = current;

previous = curr;

}

}