LeetCode 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:
Given 1->2->3->4->5->NULL,
return 1->3->5->2->4->NULL.

Nothing special.... set a flag to check whether the node is of even/odd number.

    ListNode* oddEvenList(ListNode* head) {
        if(!head || !head->next) return head;
        ListNode* oddDummy = new ListNode(0);
        ListNode* newHead = oddDummy;

        ListNode* evenDummy = new ListNode(0);
        ListNode* evenHead = evenDummy;

        ListNode* tmp = head;
        bool odd = true;
        while(tmp) {
            ListNode* next = tmp->next;
            if(odd) {
                oddDummy->next = tmp;
                oddDummy = tmp;
            } else {
                evenDummy->next  = tmp;
                evenDummy = tmp;
            }
            tmp = next;
            odd = !odd;
        }
        oddDummy->next = evenHead->next;
        evenDummy->next = NULL;
        return newHead->next;
    }