检查数组中的整数

Question

private Stack<Card> deck;//to hold cards
private Card[] protoDeck;//to hold cards before shuffling
private Random randomer;

private int cardsDealt;//how many cards used. Used for other methods
private static final int TOTALCARDS = 52;//sets limit of deck for all decks

public void shuffle(){//remove cards from deck and put back in random order
    randomer = new Random();
    int[] temp = new int[TOTALCARDS];//to keep track of random numbers
    int rand = 0;

    for (int i = 0; i < temp.length ; i++) {
        do {//keep creating randoms if 
            rand = randomer.nextInt(TOTALCARDS);
            deck.push(protoDeck[rand]);//puts the Card onto the Deck in a random position
            temp[i] = rand;
        } while (!(Arrays.asList(temp).contains(rand)));//check if the number already used  
    }
}

@PeterLawrey我稍微调整了代码如下，因为我只需要洗牌，这是一个享受，我会弹出卡片的堆栈来处理

public void shuffle() {
    randomer = new Random();
    for(int i = 0; i < TOTALCARDS; i++) {
        // pick a random card from the rest of the deck
        int j = randomer.nextInt(protoDeck.length - i) + i;
        // swap cards
        Card tmp = protoDeck[i];
        protoDeck[i] = protoDeck[j];
        protoDeck[j] = tmp;
        deck.push(protoDeck[i]);
    }

}

感谢彼得和所有其他贡献者。M.

Answer 1

从…开始

private final Card[] deck;//to hold cards before shuffling
private final Random rand = new Random();

你可以做

public void shuffle() {
    // no need the shuffle the last card.
    shuffle(deck.length - 1);
}

// will leave the first N card random without duplicates.
public void shuffle(int numberOfCards) {
    for(int i = 0; i < numberOfCards; i++) {
        // pick a random card from the rest of the deck
        int j = rand.nextInt(protoDeck.length - i) + i;
        // swap cards
        Card tmp = deck[i];
        deck[i] = deck[j];
        deck[j] = tmp;
    }
}

成本为O(N)，其中N是随机卡片的数量。

想象你有一个小甲板

AS AC AD AH 2S 2C 2D 2H

你需要随机选择第一张牌，你从牌组中选择一张并交换那张牌。假设nextInt（）是5=>2c

2C | AC AD AH 2S AS 2D 2H

书桌由随机选择+未选择的卡片组成。你没有重复的，因为相同的卡会被移动。下一个随机卡是2H，它与AC交换

2C 2H | AD AH 2S AS 2D AC

2C 2H AD | AH 2S AS 2D AC

这给你三张随机的牌和其余的。同样的数组可以再次使用，因为从排序或随机甲板开始不会使结果或多或少随机。

在回答答案时，为什么这个简单的洗牌算法会产生偏颇的结果？如果有123个，可能的结果是

123
 +- 123          - swap 1 and 1 (these are positions, not numbers)
 |   +- 123      - swap 2 and 2
 |   +- 132      - swap 2 and 3
 +- 213          - swap 1 and 2
 |   +- 213      - swap 2 and 2
 |   +- 231      - swap 2 and 3
 +- 321          - swap 1 and 3
     +- 321      - swap 2 and 2
     +- 312      - swap 2 and 3

正如你所看到的，只有6种可能的结果，所有的可能性都是一样的。