Group nodes by their index:

If there is a singly linked list, based on indexes, group all nodes with odd index first followed by the nodes with even index.

Example: Given0->1->2->3->4->5->NULL,return1->3->5->0->2->4->NULL.

Notes:

1, We are talking about the node number(index) and not the value in the nodes.

2, The relative order inside both the even and odd groups should remain as it was in the input.

3, The first node is considered even, the second node odd and so on ...

If you can solve the above problem easily, you can try to solve it with one more request again:

solve the problem above in place

Note:

You can find the definition of in place in: https://en.wikipedia.org/wiki/In-place_algorithm

Code outline:

// Definition for a node.

public class ListNode {

int val;

ListNode next;

ListNode(int x) { val = x; }

}

// solution class

public class Solution {

public ListNode oddEvenList(ListNode head) {

// please add your code here

}

// Do not need to write the main function, I will test the method oddEvenList(ListNode head) by myself.

}