LeetCode – Reverse Nodes in k-Group (Java)

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

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 pre = fake;
    int i=0;
 
    ListNode p = head;
    while(p!=null){
        i++;
        if(i%k==0){
            pre = reverse(pre, p.next);
            p = pre.next;
        }else{
            p = p.next; 
        }
    }
 
    return fake.next; 
}
 
/*
 * 0->1->2->3->4->5->6
 * |           |   
 * pre        next
 *
 * after calling pre = reverse(pre, next)
 * 
 * 0->3->2->1->4->5->6
 *          |  |
 *          pre next 
 */
public ListNode reverse(ListNode pre, ListNode next){
    ListNode last = pre.next;
    ListNode curr = last.next;
 
    while(curr != next){
        last.next = curr.next;
        curr.next = pre.next;
        pre.next = curr;
        curr = last.next;
    }
 
    return last; 
}
Category >> Algorithms  
If you want someone to read your code, please put the code inside <pre><code> and </code></pre> tags. For example:
<pre><code> 
String foo = "bar";
</code></pre>
  • SG

    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;
    }
    }