Bloomberg

Capital One

Microsoft

2020-05-16

## 328. Odd Even Linked List

### Question:

Given a singly linked list, group all odd nodes together followed by the even nodes. Please note here we are talking about the node number and not the value in the nodes.

You should try to do it in place. The program should run in O(1) space complexity and O(nodes) time complexity.

#### Example 1:

``````Input: 1->2->3->4->5->NULL
Output: 1->3->5->2->4->NULL
``````

#### Example 2:

``````Input: 2->1->3->5->6->4->7->NULL
Output: 2->3->6->7->1->5->4->NULL
``````

Note:

• The relative order inside both the even and odd groups should remain as it was in the input.
• The first node is considered odd, the second node even and so on …

### Solution:

Seperate the list into two. One is starting with odd node, another is starting with even. Then combine them together.

``````/**
* 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 oddEvenList(ListNode head) {

// Curr Node
ListNode curr = head;

// Dummy Node
ListNode oddhead = new ListNode(-1);
ListNode odd = oddhead;

// Dummy Node
ListNode evenhead = new ListNode(-1);
ListNode even = evenhead;

// Seprate the Node based on the counter
int counter = 1;
while(curr != null) {
if (counter % 2 == 1) {
odd.next = curr;
odd = curr;
} else {
even.next = curr;
even = curr;
}
curr = curr.next;
counter++;
}

// Avoid Cycle
even.next = null;

}
}
``````
``````/**
* 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 oddEvenList(ListNode head) {

ListNode odd = head, even = head.next;
ListNode evenDummy = even;
while(even != null && even.next != null){
odd.next = even.next;
odd = odd.next;
even.next = even.next.next;
even = even.next;
}
odd.next = evenDummy;