枚举每五张牌的组合

发布于 2024-08-23 13:39:14 字数 397 浏览 6 评论 0原文

我刚刚遇到了一个脑块，我有一个 Deck 对象，并且想以迭代方式从中获取每 5 张卡牌组合。有人可以告诉我如何做到这一点，我想它会是：

for(int i =0; i <52; i++){
    for(int j = i + 1 ; j < 52; j++){
        for(int k = j + 1; k < 52; k++{ 
            for(int l = k + 1; l < 52; l++){
                for(int m = l + 1; m < 52; m++){
                }
             }
         }
     }
  }

这是正确的吗？

谢谢

原文

I have just had a brain block, I have a Deck object and want to get every 5 card combination from it in a iterative manner. Could someone show me how to do this, I would imagine it would be:

for(int i =0; i <52; i++){
    for(int j = i + 1 ; j < 52; j++){
        for(int k = j + 1; k < 52; k++{ 
            for(int l = k + 1; l < 52; l++){
                for(int m = l + 1; m < 52; m++){
                }
             }
         }
     }
  }

Is this correct?

Thanks

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难如初 2024-08-30 13:39:14

是的，效果很好。如果你想枚举所有n张卡的组合，这是行不通的。

为此，您需要递归。将0张牌放入槽0中。递归枚举剩余n-1个槽中的所有n-1手牌（不包括0）。重复，将卡 1 插入插槽 0。非常简单。

编辑：一些代码：

private static final int NUM_CARDS = 52;

public static void main(String[] args) {
  enumerateAllHands(Integer.parseInt(args[0]));
}

private static void enumerateAllHands(int n) {
  if (n > NUM_CARDS) {
    throw new IllegalArgumentException();
  }
  int[] cards = new int[n];
  BitSet cardsUsed = new BitSet();
  enumerate(cards, 0, cardsUsed);
}

private static void enumerate(int[] cards, int from, BitSet cardsUsed) {
  if (from == cards.length) {
    emit(cards);
  } else {
    for (int i = 0; i < NUM_CARDS; i++) {
      if (!cardsUsed.get(i)) {
        cards[from] = i;
        cardsUsed.set(i);
        enumerate(cards, from + 1, cardsUsed);
        cardsUsed.clear(i);
      }
    }
  }
}

private static void emit(int[] cards) {
  System.out.println(Arrays.toString(cards));
}

Yes, that works fine. If you want to enumerate all n-card combinations, this doesn't work.

For that, you'd need recursion. Put card 0 in slot 0. Recursively enumerate all n-1 card hands (excluding 0) in the remaining n-1 slots. Repeat, with card 1 in slot 0. Pretty easy.

EDIT: some code:

private static final int NUM_CARDS = 52;

public static void main(String[] args) {
  enumerateAllHands(Integer.parseInt(args[0]));
}

private static void enumerateAllHands(int n) {
  if (n > NUM_CARDS) {
    throw new IllegalArgumentException();
  }
  int[] cards = new int[n];
  BitSet cardsUsed = new BitSet();
  enumerate(cards, 0, cardsUsed);
}

private static void enumerate(int[] cards, int from, BitSet cardsUsed) {
  if (from == cards.length) {
    emit(cards);
  } else {
    for (int i = 0; i < NUM_CARDS; i++) {
      if (!cardsUsed.get(i)) {
        cards[from] = i;
        cardsUsed.set(i);
        enumerate(cards, from + 1, cardsUsed);
        cardsUsed.clear(i);
      }
    }
  }
}

private static void emit(int[] cards) {
  System.out.println(Arrays.toString(cards));
}

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