題目如下：

In a deck of cards, every card has a unique integer.  You can order the deck in any order you want.

Initially, all the cards start face down (unrevealed) in one deck.

Now, you do the following steps repeatedly, until all cards are revealed:

1. Take the top card of the deck, reveal it, and take it out of the deck.
2. If there are still cards in the deck, put the next top card of the deck at the bottom of the deck.
3. If there are still unrevealed cards, go back to step 1.  Otherwise, stop.

Return an ordering of the deck that would reveal the cards in increasing order.

The first entry in the answer is considered to be the top of the deck.

 

Example 1:

Input: [17,13,11,2,3,5,7]
Output: [2,13,3,11,5,17,7]
Explanation: 
We get the deck in the order [17,13,11,2,3,5,7] (this order doesn't matter), and reorder it.
After reordering, the deck starts as [2,13,3,11,5,17,7], where 2 is the top of the deck.
We reveal 2, and move 13 to the bottom.  The deck is now [3,11,5,17,7,13].
We reveal 3, and move 11 to the bottom.  The deck is now [5,17,7,13,11].
We reveal 5, and move 17 to the bottom.  The deck is now [7,13,11,17].
We reveal 7, and move 13 to the bottom.  The deck is now [11,17,13].
We reveal 11, and move 17 to the bottom.  The deck is now [13,17].
We reveal 13, and move 17 to the bottom.  The deck is now [17].
We reveal 17.
Since all the cards revealed are in increasing order, the answer is correct.
 

 

Note:

1. 1 <= A.length <= 1000
2. 1 <= A[i] <= 10^6
3. A[i] != A[j] for all i != j

解題思路：這個題目有點意思。我們可以用輸出來反推輸入，輸出可以看成是一個佇列,記為queue，最後入隊的是最大值，最先入隊的是最小值。從輸入到輸出的邏輯是先把deck的第一個元素放入queue，然後第二個元素放入deck的最後；那麼從輸出反推的輸入的邏輯就是先把deck的最後一個元素放入到deck的第一個位置，然後再把queue的最後一個元素放入deck。

程式碼如下：

class Solution(object):
    def deckRevealedIncreasing(self, deck):
        """
        :type deck: List[int]
        :rtype: List[int]
        """
        deck.sort(reverse = True)
        res = []
        while len(deck) > 0:
            v = deck.pop(0)
            if len(res) == 0:
                res.append(v)
            else:
                res.insert(0,res.pop(-1))
                res.insert(0,v)
        return res