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最新推荐文章于 2023-08-25 16:45:50 发布

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本文介绍了解决LeetCode 950题目的方法，该题目要求玩家重新排列一副牌，使得在特定的翻牌规则下，牌面数字能够按递增顺序显示。文章提供了一种解决方案，通过逆向思考翻牌过程，实现了正确的牌组排序。

Algorithm：Leetcode 950. Reveal Cards In Increasing Order

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:

Take the top card of the deck, reveal it, and take it out of the deck.
If there are still cards in the deck, put the next top card of the deck at the bottom of the deck.
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 <= A.length <= 1000
1 <= A[i] <= 10^6
A[i] != A[j] for all i != j

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/reveal-cards-in-increasing-order
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

题目的意思是，给你一副牌，你任意排序，排好序后牌面朝下执行如下操作：

将最上面的牌翻过来，并从这一副牌里拿出来；
翻了一张牌后，将最上面的一张牌放到最底下；
重复上面两步，直到所有牌都翻过来了。
要求按照牌被翻出来的顺序，牌面数字正好是递增的。

解法一：最容易想到的可能就是把翻牌的动作倒过来执行，即：
1 从最底下拿一张牌到最上面；
2 回收一张翻出来的牌，放在最上面；
3 重复上面两步，直到所有牌都回收完毕。
按照上面的步骤：

动作	牌堆里的牌
将一张空牌放到最上面
回收17	17
将17放到最上面	17
回收13	13,17
将17放到最上面	17,13
回收11	11,17,13
将13放到最上面	13，11，17
回收7	7，13，11，17
将17放到最上面	17，7，13，11
回收5	5，17，7，13，11
将11放到最上面	11，5，17，7，13
回收3	3，11，5，17，7，13
将13放到最上面	13，3，11，5，17，7
回收2	2，13，3，11，5，17，7

这就类似一个排序的过程。先初始化一个已排序列表，每次将已排序列表的最后一个数字移动到已排序列表最前面，然后找到未排序的数字中最大的，将其插入到已排序列表的最前面。相关代码如下：

class Solution {
    public int[] deckRevealedIncreasing(int[] deck) {
        if (deck.length < 2) {
            return deck;
        }

        for (int i=deck.length; i>1; i--) {
            int maxIndex = 0;
            for (int j=1; j<i; j++) {
                if (deck[j] > deck[maxIndex]) {
                    maxIndex = j;
                }
            }
            swap(deck, maxIndex, i-1);
            insertLastElementToIndex(deck, i-1);
        }
        return deck;
    }

    private void swap(int[] deck, int i, int j) {
        int temp = deck[i];
        deck[i] = deck[j];
        deck[j] = temp;
    }

    private void insertLastElementToIndex(int[] deck, int index) {
        if (index >= deck.length-1) {
            return;
        }
        int last = deck[deck.length-1];
        System.arraycopy(deck, index, deck, index + 1, deck.length - 1 - index);
        deck[index] = last;
    }
}

Tips: 将自己的电脑加入服务器信任列表实现免密登陆

cat ~/.ssh/id_rsa.pub | ssh -p 22   root@<serverIp>    "mkdir -p ~/.ssh && cat >>  ~/.ssh/authorized_keys"