odd-even-linkedlist
Problem
Problem Description
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:
1. The relative order inside both the even and odd groups should remain as it was in the input.
2. The first node is considered odd, the second node even and so on ...
Solution
This problem is intuitive, gather all odd nodes together, and all even nodes together. then odd list point to even node list.
List Node is a bit not intuitive to implement, so you need to understand pointer for list node. Will using an example below to demonstrate how pointer works and how to gather all odd and even nodes.
will need 3 new heads,
odd -- point to odd nodes, while iterate through list nodes.
even -- point to even nodes
evenHead -- need to remember the start position for even head.
init,
odd = head
andeven = head.next
andevenHead = even
After init, traverse through list nodes, here head and evenHead will remain, not move, instead, odd and even will move to corresponding position.
each traverse,
odd = odd.next.next
(by pass even node), andeven = even.next.next
(by passing odd node)until
even == null || even.next == null
, done scanning all nodes, evenHead is in the even list nodes head.point
odd.next = evenHead
.now head is the result.
For example:

Complexity Analysis
Time Complexity: O(N)
Space Complexity: O(1)
N - the number of nodes in linked list
Code
class Solution {
/**
* 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) {
if (head == null) return head;
ListNode odd = head;
ListNode even = head.next;
ListNode evenHead = even;
while (even != null && even.next != null) {
odd.next = odd.next.next;
even.next = even.next.next;
odd = odd.next;
even = even.next;
}
odd.next = evenHead;
return head;
}
}
}
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