解决思路

链表

用两个链表,一个链表放小于x的节点,一个链表放大于等于x的节点
最后,拼接这两个链表.

  public ListNode partition(ListNode head, int x) {
        ListNode dummy1 = new ListNode(-1);
        ListNode dummy2 = new ListNode(-1);
        ListNode p1 = dummy1;
        ListNode p2 = dummy2;
        while (head != null) {
            if (head.val < x) {
                p1.next = head;
                p1 = p1.next;
            } else {
                p2.next = head;
                p2 = p2.next;
            }
            head = head.next;
        }
        p1.next = dummy2.next;
        p2.next = null;
        return dummy1.next; 
    }