Question

我正在尝试编写一个非常简单的扑克游戏。我只是使用非面部卡，2-9，没有套件或任何类似的东西。我正在试图弄清楚如何编写一种方法来确定五张牌是否是一个满屋，这是一对和一种3。我有用户输入5个整数，表示卡值并将它们存储在一个数组中。我试着写这样的东西：

public static boolean containsFullHouse(int[] hand)
{
     for (int i = 0; i < hand.length; i++){
         int count = 0;

         for (int j = 0; j < hand.length; j++){
              if (hand[i] == hand[j]){
                 count++;}

              if (count == 3){
                 return true;}
         }
     }
         for(int i = 0; i < hand.length; i++){
              for(int j = i + 1; j < hand.length; j++){ 
                  if(hand[i] == hand[j]){
                        return true;}
                   }
               }
         }  
    return false;
}

Answer 1

您需要计算每个数字的出现次数，并创建所谓的基数映射。然后基数必须是（2,3）或（3,2）。如果不使用guava或Apache Commons Collections（其中包含方便的方法），可以通过以下方式完成：

public static boolean isFullHouse(final int[] input) {
    if (input.length != 5) { throw new IllegalArgumentException("need 5 integers"); }
    Map<Integer, Integer> cardinalityMap = new HashMap<>();
    for (int i : input) {
        if (cardinalityMap.containsKey(i)) {
            cardinalityMap.put(i, cardinalityMap.get(i) + 1);
        }
        else {
            cardinalityMap.put(i, 1);
        }
    }

    if (cardinalityMap.size() != 2) { return false; }
    Collection<Integer> occurences = cardinalityMap.values();
    int first = occurences.iterator().next();
    return first == 2 || first == 3;
}

Answer 2

<强>问题：

您正在检查索引i两次，虽然正确（因为您检查count == 3），但这是不必要的。
在检查其他2之前，你也会回来。
第二个循环将返回true，因为它将找到前一循环中的数字。

如果您对它们进行排序，您只需检查两侧的两对卡是否相同，并检查中间卡是否与其中任何一张相同。所以像这样：

Arrays.sort(hand);
return (hand[0] == hand[1] && hand[3] == hand[4] &&
        (hand[2] == hand[1] || hand[2] == hand[3]));

或者，如果您想修复功能：

public static boolean containsFullHouse(int[] hand)
{
     // a variable that keeps track of one of the 3-of-a-kind indices (used in 2-of-a-kind check)
     int pos = -1;

     for (int i = 0; i < hand.length && pos == -1; i++){
         // start count at one instead
         int count = 1;

         // start j from next position rather than 0
         for (int j = i+1; j < hand.length && pos == -1; j++){
              if (hand[i] == hand[j]) {
                 count++;
              }

              if (count == 3) {
                 pos = i;
              }
         }
     }

     // if we didn't find 3-of-a-kind, return false
     if (pos == -1)
         return false;

     // look for 2-of-a-kind
     for(int i = 0; i < hand.length; i++){
         // exclude elements that match one of the 3-of-a-kind
         if (hand[i] != hand[pos]){
             for(int j = i + 1; j < hand.length; j++){ 
                 if(hand[i] == hand[j]){
                        return true;
                 }
             }
         }
     }

    return false;
}

Answer 3

我会使用Apache Commons的CollectionUtils.getCardinalityMap来做这个

public static void main(String[] args) {

    Integer[] fullHouse = new Integer[]{7, 7, 7, 4, 4};
    Integer[] notFullHouse = new Integer[]{2, 2, 2, 2, 3};
    Integer[] notFullHouse2 = new Integer[]{1, 4, 2, 2, 3};

    System.out.println(isFullHouse(fullHouse));
    System.out.println(isFullHouse(notFullHouse));
    System.out.println(isFullHouse(notFullHouse2));

}

private static boolean isFullHouse(Integer[] cards){
    Map<Integer,Integer> cardinalityMap = CollectionUtils.getCardinalityMap(Arrays.asList(cards));

    if(cardinalityMap.size() == 2) {
        if (cardinalityMap.values().containsAll(Arrays.asList(2, 3))) {
            return true;
        }
        return false;
    }
    return false;
}

Answer 4

满屋由2个不同的整数组成，因此请为两者保留一个计数器。它还需要跟踪2个不同的值。如果你把它结合起来就可以得到这样的结果：

public static boolean containsFullHouse(int[] hand)
{
    int value1 = -1, count1 = 0;
    int value2 = -1, count2 = 0;

    for (int i = 0; i < hand.length; i++) {
        if(hand[i] == value1) {
            // Found another value1 card
            count1++;
        } else if(hand[i] == value2) {
            // Found another value2 card
            count2++;
        } else if(value1 == -1) {
            // Found a new card, store as value1
            value1 = hand[i];
            count1++;
        } else if(value2 == -1) {
            // Found a new card, store as value2
            value2 = hand[i];
            count2++;
        } else {
            // Found a third card, so it cannot be a full house!
            return false;
        }
    }
    if(value2 == -1) {
        // Found 'five of a kind'?!
        return false;
    }

     // Check if it is a full house
    return (count1 == 3 && count2 == 2) || (count1 == 2 && count2 == 3;)
}

如何确定我输入数组的5个数字是否是使用Java顺序的？

4 个答案: