328. Odd Even Linked List

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 ...

One pass

public ListNode oddEvenList(ListNode head) {
    if(head == null || head.next == null || head.next.next == null) return head;
    ListNode odd = head, prev = head.next, curr = head.next.next;
    // prev always even, curr always odd
    // move curr behind the odd
    while(curr != null) {
        prev.next = curr.next;
        curr.next = odd.next;
        odd.next = curr;
        odd = odd.next;
        prev = prev.next;
        if(prev == null) break;
        curr = prev.next;
    }
    return head;
}

Move odd to left for one step in a single traverse O(N2)

This approach is a trash…

public ListNode oddEvenList(ListNode head) {
    if(head == null) return null;
    ListNode curr = head, prev = null;
    int len = 1;
    while(curr.next != null) {
        prev = curr;
        curr = curr.next;
        len++;
    }
    ListNode tail = null;
    if(len % 2 == 1) {
        len--;
        if(len > 1) {
            tail = curr;
            prev.next = null;
        }
    }
    curr = head;
    ListNode n = null;
    while(len >= 4) {
        int i = len - 2;
        prev = curr;
        n = prev.next;
        while(i > 0) {
            prev.next = n.next;
            n.next = n.next.next;
            prev.next.next = n;
            prev = prev.next.next;
            n = prev.next;
            i -= 2;
        }
        curr = curr.next;
        len -= 2;
    }
    if(tail != null) {
        tail.next = curr.next;
        curr.next = tail;
    }
    return head;
}