题解 | #链表的奇偶重排#

/**
     * so easy
     * @param head
     * @return
     */
    public static ListNode oddEvenList(ListNode head) {
        // write code here
        if (head == null) {
            return null;
        }
        ListNode t1 = head;
        ListNode t2 = head.next;
        ListNode tail = t1;
        ListNode cur = head;
        ListNode old = head.next;
        while (old != null) {
            ListNode node = old.next;
            if (node == null) {
                break;
            }
            ListNode node1 = node.next;
            cur.next = node;
            cur = node;
            old.next = node1;
            old = node1;
            tail = node;
        }
        tail.next = t2;
        return t1;
    }