Question

我有一个必须执行以下操作的方法：

for (int a01 = 1; a01 <= 25; a01++) {
    for (int a02 = a01 + 1; a02 <= 25; a02++) {
        for (int a03 = a02 + 1; a03 <= 25; a03++) {
            ...
            System.out.println(a01 + "," + a02 + "," + ... + "," + a015);
        }
    }
}

我想指定嵌套的数量（在上面的例子中，我想要15嵌套）。有没有办法在这里使用递归编程？

Answer 1

是。这可以通过递归编程来执行。

我假设你不喜欢在源代码中嵌入这些嵌套代码 - 就像在你的例子中一样，因为这是非常难看的编程 - 就像评论员解释的那样。

以下（伪类Java）代码说明了它。我假设嵌套的固定深度。那么你实际上喜欢循环一个维度深度的整数向量。

int[] length = new int[depth];
int[] counters = new int[depth];

数组counters必须初始化为0（Arrays.fill(counters,0)）。必须将数组length初始化为相应for循环的迭代次数。

我假设你喜欢在内循环中执行某个操作。我会打电话给你 performOperation(int[] counters); - 它取决于多维计数器，即外部的计数器。

然后你可以通过调用

来运行嵌套的for循环

nestedLoopOperation(counters, length, 0);

，其中

void nestedLoopOperation(int[] counters, int[] length, int level) {
    if(level == counters.length) performOperation(counters);
    else {
        for (counters[level] = 0; counters[level] < length[level]; counters[level]++) {
            nestedLoopOperation(counters, length, level + 1);
        }
    }
}

在您的情况下，您的System.out.println（）将是

performOperation(int[] counters) {
    String counterAsString = "";
    for (int level = 0; level < counters.length; level++) {
        counterAsString = counterAsString + counters[level];
        if (level < counters.length - 1) counterAsString = counterAsString + ",";
   }
   System.out.println(counterAsString);
}

Answer 2

我创建了这个程序，以显示所有不同的卡片组合（非重复）。它使用递归for循环。也许它可以帮助你。

//I'm lazy, so yeah, I made this import...
import static java.lang.System.out;

class ListCombinations {

    //Array containing the values of the cards
    static Symbol[] cardValues = Symbol.values();

    //Array to represent the positions of the cards,
    //they will hold different card values as the program executes
    static Symbol[] positions = new Symbol[cardValues.length];

    //A simple counter to show the number of combinations
    static int counter = 1;

    /*Names of cards to combine, add as many as you want, but be careful, we're
    talking about factorials here, so 4 cards = 24 different combinations (4! = 24),
    but 8 cards = 40320 combinations and 13 cards = 6.23 billion combinations!*/
    enum Symbol {
        AofSpades, TwoofSpades, ThreeofSpades, FourofSpades
    }

    public static void main(String args[]) {

        //I send an argument of 0 because that is the first location which
        //we want to add value to. Every recursive call will then add +1 to the argument.
        combinations(0);
    }

    static void combinations(int locNumber) {

        /* I use a recursive (repeating itself) method, since nesting for loops inside
        * each other looks nasty and also requires one to know how many cards we will
        * combine. I used 4 cards so we could nest 4 for loops one after another, but
        * as I said, that's nasty programming. And if you add more cards, you would
        * need to nest more for loops. Using a recursive method looks better, and gives
        * you the freedom to combine as many cards as you want without changing code. */

        //Recursive for loop nesting to iterate through all possible card combinations
        for(int valueNumber = 0; valueNumber < cardValues.length; valueNumber++) {
            positions[locNumber] = cardValues[valueNumber];
            if (locNumber < (cardValues.length-1)) {
                combinations(locNumber + 1);
            }

            //This if statement grabs and displays card combinations in which no card value
            // is repeated in the current "positions" array. Since in a single deck,
            // there are no repeated cards. It also appends the combination count at the end.
            if (locNumber == (cardValues.length-1) && repeatedCards(positions)) {
                for (int i = 0; i < cardValues.length; i++) {
                out.print(positions[i]);
                out.print(" ");
                }
                out.printf("%s", counter);
                counter++;
                out.println();
            }
        }
    }

    static boolean repeatedCards(Symbol[] cards) {

        /*Method used to check if any cards are repeated in the current "positions" array*/

        boolean booleanValue = true;

        for(int i = 0; i < cardValues.length; i++) {
            for(int j = 0; j < cardValues.length; j++) {
                if(i != j && cards[i] == cards[j]) {
                    booleanValue = false;
                }
            }
        }
        return booleanValue;
    }
}

如何递归地嵌套for循环？

2 个答案: