LeetCode之Reveal Cards In Increasing Order（Kotlin）

嘟囔發表於2019-01-01
LeetCodeKotlin

問題： 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.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.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
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方法： 建立第一個陣列對輸入進行排序作為取出的牌，然後建立第二個陣列作為牌桌，陣列第一個元素表示bottom，最後一個元素表示top；然後重複“把bottom的牌放到top”和“把取出的牌放回top”這兩個操作，直到所有取出的牌被放回牌桌，則最終牌桌上的牌反向輸出即為答案（牌桌上的牌是從bottom到top）。

具體實現：

class RevealCardsInIncreasingOrder {
    fun deckRevealedIncreasing(deck: IntArray): IntArray {
        val sortArray = mutableListOf<Int>()
        sortArray.addAll(deck.toList())
        sortArray.sort()
        val result = mutableListOf<Int>()
        for (index in sortArray.lastIndex downTo 0) {
            if (result.isNotEmpty()) {
                result.add(result.lastIndex, result.removeAt(0))
            }
            result.add(sortArray[index])
        }
        result.reverse()
        return result.toIntArray()
    }
}

fun main(args: Array<String>) {
    val input = intArrayOf(17,13,11,2,3,5,7)
    val revealCardsInIncreasingOrder = RevealCardsInIncreasingOrder()
    CommonUtils.printArray(revealCardsInIncreasingOrder.deckRevealedIncreasing(input).toTypedArray())
}
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有問題隨時溝通

具體程式碼實現可以參考Github

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