题解 | #链表的奇偶重排#

O(n),O(1)

    public ListNode oddEvenList (ListNode head) {
        // write code here
        if (head == null || head.next == null) return head;
        ListNode h1 = head, h2 = head.next;
        ListNode p1 = h1, p2 = h2;
        // 你追我赶
        while (p1.next != null && p2.next != null){
            if (p2.next != null){
                p1.next = p2.next;
                p1 = p1.next;
            }
            if (p1.next != null){
                p2.next = p1.next;
                p2 = p2.next;
            }
        }
        // 处理最后一个节点的next指针，否则可能会造成环形链表
        p2.next = null;
        p1.next = h2;
        return h1;
    }