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

Maintain two pointers to swap the next pointers for even and odd index. Iterate over the linked list and change the same.

###### Code

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class Solution { public ListNode oddEvenList(ListNode head) { if(head == null) return head; ListNode result = head; ListNode p1 = head; ListNode p2 = head.next; ListNode connectNode = head.next; while(p1 != null && p2 != null){ ListNode t = p2.next; if(t == null) break; p1.next = p2.next; p1 = p1.next; p2.next = p1.next; p2 = p2.next; } p1.next = connectNode; return result; } } |

#### Complexity Analysis

Time complexity: O(N).

Space Complexity: O(1).

We encourage you to write a comment if you have a better solution or having any doubt on the above topic.