I'm working on an algorithm to rearrange a singly linked list based on node positions using only a single stack as auxiliary space. The goal is to modify the linked list such that all nodes at odd positions (1st, 3rd, 5th, etc.) appear before all nodes at even positions (2nd, 4th, 6th, etc.), while preserving the relative order within each group.

For example, given a linked list 1->2->3->4->5->6->7->8, the rearranged list should be 1->3->5->7->2->4->6->8. Here, all odd-positioned nodes (1,3,5,7) maintain their relative order and appear before all even-positioned nodes (2,4,6,8), which also maintain their order.

The key constraints that make this problem challenging are:

1. We can only use one stack as additional space (besides O(1) variables)

2. The solution should work in a single pass through the linked list

3. No other data structures are allowed

4. The original linked list should be modified in-place

5. Node values are arbitrary and cannot be used for position identification

I've attempted storing even-positioned nodes in the stack while linking odd-positioned nodes directly, but I'm having trouble maintaining the correct links when reconstructing the list. My current implementation loses some connections when trying to reattach the even-positioned nodes from the stack.

Can you help identify an efficient approach that satisfies these constraints? I'm particularly interested in understanding how to handle the pointer manipulation when reattaching nodes from the stack while preserving the required ordering.

I'm implementing a custom linked list rearrangement algorithm in Java that needs to reorder nodes based on their positions. The challenge is to move all odd-positioned nodes (1st, 3rd, 5th...) before all even-positioned nodes (2nd, 4th, 6th...), maintaining relative order within each group, using only a single stack as auxiliary space.

Here's my attempted implementation:

class ListNode {
    int val;
    ListNode next;
    
    ListNode(int val) {
        this.val = val;
        this.next = null;
    }
}

public class LinkedListRearranger {
    public ListNode rearrangeByPosition(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }
        
        Stack<ListNode> evenNodes = new Stack<>();
        ListNode current = head;
        ListNode oddTail = null;
        int position = 1;
        
        // First pass: separate odd and even positioned nodes
        while (current != null) {
            ListNode nextNode = current.next;
            current.next = null;  // Break existing link
            
            if (position % 2 == 1) {
                // Handle odd-positioned nodes
                if (oddTail == null) {
                    oddTail = current;
                    head = current;  // Keep track of new head
                } else {
                    oddTail.next = current;
                    oddTail = current;
                }
            } else {
                // Push even-positioned nodes to stack
                evenNodes.push(current);
            }
            
            current = nextNode;
            position++;
        }
        
        // Second pass: reconnect even nodes from stack
        while (!evenNodes.isEmpty()) {
            oddTail.next = evenNodes.pop();
            oddTail = oddTail.next;
        }
        
        return head;
    }
}

Expected Behavior: For the input list: 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 Expected output: 1 -> 3 -> 5 -> 7 -> 2 -> 4 -> 6 -> 8

The algorithm should:

1. Process the list in a single traversal, using O(n/2) stack space for even nodes

2. Maintain the relative ordering of odd-positioned nodes (1,3,5,7)

3. Maintain the relative ordering of even-positioned nodes (2,4,6,8)

4. Properly handle edge cases like lists with odd lengths

5. Avoid creating any cycles in the final list

Actual Behavior: My implementation is encountering several issues:

1. Some nodes are getting lost during the rearrangement

2. The final list sometimes contains incorrect connections

3. When processing lists with odd lengths, the last node isn't handled properly

4. In some cases, the list ends up with cycles

Test case that fails:

ListNode head = new ListNode(1);

head.next = new ListNode(2);

head.next.next = new ListNode(3);

head.next.next.next = new ListNode(4);

head.next.next.next.next = new ListNode(5);

Expected: 1->3->5->2->4

Actual: Getting incorrect connections or lost nodes

I suspect the issues might be related to:

1. Improper handling of the next pointers when breaking and reconnecting nodes

2. Not correctly tracking the tail of the odd-positioned nodes list

3. Edge cases I haven't considered in the stack operations

I've tried several variations:

1. Using a dummy head node to simplify first node handling

2. Different approaches to breaking and reconnecting links

3. Maintaining a separate reference for the even nodes list

But each attempt either violates the space complexity requirement or fails to maintain the correct node ordering.

What am I doing wrong in my implementation? How can I fix these issues while maintaining the space complexity constraints and ensuring proper node connections?

I'm working on an algorithm to rearrange a singly linked list based on node positions using only a single stack as auxiliary space. The goal is to modify the linked list such that all nodes at odd positions (1st, 3rd, 5th, etc.) appear before all nodes at even positions (2nd, 4th, 6th, etc.), while preserving the relative order within each group.

For example, given a linked list 1->2->3->4->5->6->7->8, the rearranged list should be 1->3->5->7->2->4->6->8. Here, all odd-positioned nodes (1,3,5,7) maintain their relative order and appear before all even-positioned nodes (2,4,6,8), which also maintain their order.

The key constraints that make this problem challenging are:

1. We can only use one stack as additional space (besides O(1) variables)

2. The solution should work in a single pass through the linked list

3. No other data structures are allowed

4. The original linked list should be modified in-place

5. Node values are arbitrary and cannot be used for position identification

I've attempted storing even-positioned nodes in the stack while linking odd-positioned nodes directly, but I'm having trouble maintaining the correct links when reconstructing the list. My current implementation loses some connections when trying to reattach the even-positioned nodes from the stack.

Can you help identify an efficient approach that satisfies these constraints? I'm particularly interested in understanding how to handle the pointer manipulation when reattaching nodes from the stack while preserving the required ordering.

I'm implementing a custom linked list rearrangement algorithm in Java that needs to reorder nodes based on their positions. The challenge is to move all odd-positioned nodes (1st, 3rd, 5th...) before all even-positioned nodes (2nd, 4th, 6th...), maintaining relative order within each group, using only a single stack as auxiliary space.

Here's my attempted implementation:

class ListNode {
    int val;
    ListNode next;
    
    ListNode(int val) {
        this.val = val;
        this.next = null;
    }
}

public class LinkedListRearranger {
    public ListNode rearrangeByPosition(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }
        
        Stack<ListNode> evenNodes = new Stack<>();
        ListNode current = head;
        ListNode oddTail = null;
        int position = 1;
        
        // First pass: separate odd and even positioned nodes
        while (current != null) {
            ListNode nextNode = current.next;
            current.next = null;  // Break existing link
            
            if (position % 2 == 1) {
                // Handle odd-positioned nodes
                if (oddTail == null) {
                    oddTail = current;
                    head = current;  // Keep track of new head
                } else {
                    oddTail.next = current;
                    oddTail = current;
                }
            } else {
                // Push even-positioned nodes to stack
                evenNodes.push(current);
            }
            
            current = nextNode;
            position++;
        }
        
        // Second pass: reconnect even nodes from stack
        while (!evenNodes.isEmpty()) {
            oddTail.next = evenNodes.pop();
            oddTail = oddTail.next;
        }
        
        return head;
    }
}

Expected Behavior: For the input list: 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 Expected output: 1 -> 3 -> 5 -> 7 -> 2 -> 4 -> 6 -> 8

The algorithm should:

1. Process the list in a single traversal, using O(n/2) stack space for even nodes

2. Maintain the relative ordering of odd-positioned nodes (1,3,5,7)

3. Maintain the relative ordering of even-positioned nodes (2,4,6,8)

4. Properly handle edge cases like lists with odd lengths

5. Avoid creating any cycles in the final list

Actual Behavior: My implementation is encountering several issues:

1. Some nodes are getting lost during the rearrangement

2. The final list sometimes contains incorrect connections

3. When processing lists with odd lengths, the last node isn't handled properly

4. In some cases, the list ends up with cycles

Test case that fails:

ListNode head = new ListNode(1);

head.next = new ListNode(2);

head.next.next = new ListNode(3);

head.next.next.next = new ListNode(4);

head.next.next.next.next = new ListNode(5);

Expected: 1->3->5->2->4

Actual: Getting incorrect connections or lost nodes

I suspect the issues might be related to:

1. Improper handling of the next pointers when breaking and reconnecting nodes

2. Not correctly tracking the tail of the odd-positioned nodes list

3. Edge cases I haven't considered in the stack operations

I've tried several variations:

1. Using a dummy head node to simplify first node handling

2. Different approaches to breaking and reconnecting links

3. Maintaining a separate reference for the even nodes list

But each attempt either violates the space complexity requirement or fails to maintain the correct node ordering.

What am I doing wrong in my implementation? How can I fix these issues while maintaining the space complexity constraints and ensuring proper node connections?

• java
• algorithm
• data-structures
• linked-list
• stack

• 2 That seems odd. Why would anyone use a stack? Just use a few node variables... – no comment Commented 18 hours ago
• "Expected: 1->3->5->2->4 Actual: Getting incorrect connections or lost nodes": the code will run deterministically. You are not getting lost nodes for this case. If you believe you have cases where nodes are lost, please provide the input for which that happens, so we can reproduce it. If not, please only report the problem that you actually have with the given examples. – trincot Commented 16 hours ago
• @nocomment - I agree. If a stack is used, the list has to be traversed forwards to build the stack, then traversed backwards to pop nodes from the stack. It would be simpler to split list into two lists: even nodes, odd nodes, then append the second list to the first list. If there is an odd number of nodes, the last nodes next reference needs to be set to null. – rcgldr Commented 2 hours ago

3 Answers 3

You wrote that "nodes are lost", but for the example you have provided this does not happen. The problem that prevails is that the order in which the even-positioned nodes are appended is reversed. This is because a stack is a first-in-last-out data structure.

You could fix this order issue by rewriting the second phase so the popped nodes are prefixed to the second partition instead of being postfixed to it:

        ListNode firstEven = null;
        while (!evenNodes.isEmpty()) {
            current = evenNodes.pop();
            current.next = firstEven;
            firstEven = current;
        }
        oddTail.next = firstEven;

This will solve the issue.

But it is really not necessary to use a stack here: Just like you can rewire nodes to only include the nodes that were at odd positions, in the same way you can (at the same time) rewire the other (even positioned) nodes to stick together too. No other data structure is needed.

For instance, it can look like this:

    static public ListNode rearrangeByPosition(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }
        
        ListNode evenHead = head.next;
        ListNode oddTail = head;
        ListNode evenTail;
        ListNode current = evenHead;

        // Single pass: separate odd and even positioned nodes
        while (current != null) {
            // current's (original) position is even, append its successor to the odd partition
            evenTail = current;
            current = oddTail.next = current.next;
            if (current == null) break;
            // current's (original) position is odd, append its successor to the even partition
            oddTail = current;
            current = evenTail.next = current.next;
        }
        oddTail.next = evenHead; // Link the two partitions
        return head;
    }

Here the loop takes care of two steps at a time, so there is no need for the position variable. Note the symmetry between the first half and the second half of the loop body. After the loop you have the nodes group so that they are in the linked list defined by head (the odd positioned nodes) or the linked list defined by evenHead. All that remains is to link the tail node of the first linked list to the head node of the second one, which is what the statement after the loop accomplishes.

Perhaps this approach can serve as a starting point for you.

public class LinkedListRearranger {

   public ListNode rearrangeByPositionA( ListNode head ) {
      ListNode even = head.next;
      ListNode auxA = head;
      ListNode auxB = even;
      for( int i = 1; true; i ++ ) {
           // alternatively we add to each branch (even and odd) the next 
           // node of the opposite branch, if it does not exist, we 
           // cancel the reference to ‘next’ and exit the loop.
         if( i % 2 == 0 ) {
            if( auxA.next == null ) {
               auxB.next = null;
               break;
            }
            auxB.next = auxA.next;
            auxB = auxB.next;
         }
         else {
            if( auxB.next == null ) {
               auxA.next = null;
               break;
            }
            auxA.next = auxB.next;
            auxA = auxA.next;
         }
      }
        // we join the two sections
      auxA.next = even;
      show( head );
      return head;
   }

   void show( ListNode node ) {
      System.out.println( " -- show --" );
      while( true ) {
         System.out.print( node.val + " " );
         if( node.next != null ) {
            node = node.next;
         }
         else {
            return;
         }
      }
   }

   void ini() {
      ListNode head = new ListNode( 1 );
      ListNode aux = head;
      for( int i = 1; i < 8; i ++ ) {
         aux.next = new ListNode( i + 1 );
         aux = aux.next;
      }
      show( head );
      System.out.println( "-- --" );
      rearrangeByPosition( head );
   }

   public static void main( String[] args ) {
      new LinkedListRearranger().ini();
   }
}

public ListNode rearrangeByPosition(ListNode head) { if (head == null || head.next == null) return head;

Stack<ListNode> evenNodes = new Stack<>();
ListNode current = head;
ListNode oddTail = null;
int position = 1;

// Separate odd and even nodes
while (current != null) {
    ListNode nextNode = current.next;
    if (position % 2 == 1) {
        if (oddTail == null) {
            oddTail = current;
            head = current; // New head
        } else {
            oddTail.next = current;
            oddTail = current;
        }
    } else {
        evenNodes.push(current);
    }
    current = nextNode;
    position++;
}

// Reconnect even nodes
while (!evenNodes.isEmpty()) {
    oddTail.next = evenNodes.pop();
    oddTail = oddTail.next;
}

oddTail.next = null; // Avoid cycles
return head;

}