用 Java 洗一副纸牌
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/39557701/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Shuffle a deck of cards in Java
提问by Illy P.
I've read many things on the topic, but I am still having misgivings about it. Please, help!
我已经阅读了很多关于该主题的内容，但我仍然对此有疑虑。请帮忙！
I create a deck of 52 cards:
我创建了一副 52 张牌：
int[] deck = new int[52];
for (int i = 0; i < deck.length; i++) {deck[i] = i;}
Now, I'd like to shuffle the deck properly, not by following any pattern or function.
现在，我想正确洗牌，而不是遵循任何模式或功能。
I am going to use the Math.random()method in my algorithm:
我将在我的算法中使用Math.random()方法：
for (int i = 0; i < deck.length; i++) {
int j = (int)(Math.random() * deck.length); // Get a random index out of 52
int temp = deck[i]; // Swap the cards
deck[i] = deck[j];
deck[j] = temp;
}
The problem I am having here, however, is that I sometimes might get duplicates; as a result, some cards/values are missing altogether.
然而，我在这里遇到的问题是我有时可能会得到重复；结果，一些卡片/值完全丢失。
I am almost certain that I am not discarding the indices/values I have already used, which is probably the core of the abovementioned problem. But how do I go about it?
我几乎可以肯定我没有丢弃我已经使用过的索引/值，这可能是上述问题的核心。但是我该怎么做呢？
Any suggestions? Possibly, following my same train of thought. Thanks a bunch!
有什么建议？可能，按照我的思路。谢谢一堆！
回答by Fayaz
Please look if this helps:
请看看这是否有帮助：
boolean init[] = new boolean[52];
Random r = new Random();
int getNextCard(){
int i = r.nextInt(52);
while(init[i])
i = r.nextInt(52);
init[i] = true;
return i;
}
void shuffleCards() {
int[] deck = new int[52];
for (int i = 0; i < deck.length; i++) {
deck[i] = getNextCard();
}
System.out.println(Arrays.toString(deck));
}
Idea is to use a Randomto generate a random number between 0 and 52. Once a number is taken, we will mark it as taken in the boolean array init.So, next time the same number comes up, we again generate the random number until we get a number we haven't already taken.
想法是用aRandom生成一个0到52之间的随机数。一旦取了一个数，我们会在布尔数组中将其标记为已取init。所以，下次出现相同的数字时，我们再次生成随机数，直到我们获取我们尚未使用的号码。
回答by felipeptcho
Simply use a new array for the shuffled cards and fill it using random cards removed from the prior array.
只需为洗牌的卡片使用一个新数组，并使用从先前数组中删除的随机卡片填充它。
import java.util.ArrayList;
public class Test {
private static final int DECK_SIZE = 52;
public static void main(String args[]) {
ArrayList<Integer> deck = new ArrayList<Integer>();
for (int i = 0; i < DECK_SIZE; ++i) {
deck.add(i);
}
ArrayList<Integer> shuffledDeck = new ArrayList<Integer>();
while (deck.size() > 0) {
int index = (int) (Math.random() * deck.size());
shuffledDeck.add(deck.remove(index));
}
System.out.println(shuffledDeck.toString());
}
}
Or just use Collections.shuffle():
或者只是使用Collections.shuffle()：
import java.util.ArrayList;
import java.util.Collections;
public class Test {
private static final int DECK_SIZE = 52;
public static void main(String args[]) {
ArrayList<Integer> deck = new ArrayList<Integer>();
for (int i = 0; i < DECK_SIZE; ++i) {
deck.add(i);
}
Collections.shuffle(deck);
System.out.println(deck);
}
}
