Question

嘿伙计们，我正在尝试用Java创建一个系统的列表生成器但是我遇到了轻微的错误。我们的例子是披萨。

问题：假设你有6个配料来制作比萨饼。您可以制作5种顶级披萨的多少种变种？

例如：

[意大利辣香肠，培根，菠萝，洋葱，蘑菇，辣椒] [意大利辣香肠，培根，菠萝，洋葱，蘑菇，奶酪] 等...

在开发过程中，我已经练习了3个可能的选项，分别为6个，因为我知道总共有三个组合，每个组合有两个元素，这使得我的多维数组很容易生成。

public class SystematicList {

static String options[] = {
    "Pepperoni",
    "Bacon",
    "Cheese"
};

public static void main(String[] args) {

    /**
     * Generate a multi-dimensional array to represent the combinations of
     * pizza toppings we can possibly have from the available <code>options</code>
     *
     */
    String[][] combos = new String[3][2];

    /**
     * Ideally what we would like to create.
     * combos[0][0] = "Pepperoni";
     * combos[0][1] = "Bacon";
     * combos[1][0] = "Pepperoni";
     * combos[1][1] = "Cheese";
     * combos[2][0] = "Bacon";
     * combos[2][1] = "Cheese";
     */
    for (int pizzas = 0; pizzas < combos.length; pizzas++) {
        for (int toppings = 0; toppings < combos[pizzas].length; toppings++) {
            combos[pizzas][toppings] = options[0]; // <<< issue : element.
            System.out.print(" " + combos[pizzas][toppings]);
        }
        System.out.println("");
    }

    /*
     * Current Output :
     * run:
     *  Pepperoni Bacon
     *  Pepperoni Bacon
     *  Pepperoni Bacon
     * BUILD SUCCESSFUL (total time: 0 seconds)
     * 
     * ^ this is obvious, as I currently do not know how I'll select an
     * element from the array of topping options before/after a specified
     * index [example: 0 which would than range 1 - 2 apposed to 0 - 2
     * thus 'dropping' an optional element, making this much easier.
     *
     * Loop is virtually only doing the actions of :
     * 0 : 1,2
     * 1 : 1,2
     * 2 : 1,2
     */
}

}

我决定生成一个2D列表来存储值。

int [#available combination] [＃possible values]

我当前的代码假设我们已经知道可能的组合（当然我们当然没有），并且无论我们总是可以确定值（可以构建多少5,4,3,2个顶部比萨饼）

我如何从options []中选择一个元素，同时确保没有其他数组已包含您尝试插入的元素。我会试过Arrays.contains或Arrays.equals但是我不知道要插入什么来比较，combos [pizzas-1]？

if (!Arrays.contains(combos[pizzas], combos[pizzas-1]) {
combos[pizzas][toppings] = options[?];
}

Answer 1

一种方法是仅查看尚未考虑的配料。因此，例如当从[意大利辣香肠，培根，菠萝，洋葱，蘑菇，辣椒，奶酪]中选择三种配料时，您可能决定跳过奶酪，然后选择辣椒，然后跳过蘑菇和洋葱，并选择菠萝和培根。当然，您需要留下足够的配料来完成披萨，因此您不能跳过奶酪，辣椒，蘑菇，洋葱和菠萝。 JavaScript中的示例代码使用递归算法为您提供了一个想法。

function combos(toppings, count) {
  var result = [];
  helper(result, [], toppings, toppings.length, count);
  return result;
}

function helper(result, selection, toppings, length, count) {
  if (count == 0) // no more toppings left to add
    result.push(selection); // this must be a solution
  else
    // start at --count because we need to be able to add other toppings
    for (var i = --count; i < length; i++)
      // add the selected topping to the selection
      // then consider toppings we haven't looked at yet
      helper(result, selection.concat([toppings[i]]), toppings, i, count);
}

Answer 2

我不会使用数组来保存披萨选项。您有两列浇头。如果我想要一块普通的奶酪比萨怎么办？

这是一个使用递归来构建不同数量浇头的比萨饼的解决方案。

public static List<String> getPizzas( List<String> ingredients )
{
    if( ingredients.isEmpty() )
    {
        List<String> pizzas = new ArrayList<String>();
        String plainPizza = "";
        pizzas.add( plainPizza );
        return pizzas;
    }
    else if( ingredients.size() == 1 )
    {
        List<String> pizzas = new ArrayList<String>();
        String plainPizza = "";
        String oneToppingPizza = ingredients.get( 0 );

        pizzas.add( plainPizza );
        pizzas.add( oneToppingPizza );
        return pizzas;
    }
    else
    {
        List<String> pizzas = new ArrayList<String>();

        for( String onePizza : getPizzas( ingredients.subList( 1, ingredients.size() )))
        {
            String pizzaWithoutTopping = onePizza;
            String pizzaWithTopping = ingredients.get( 0 ) + " " + onePizza;
            pizzas.add( pizzaWithTopping );
            pizzas.add( pizzaWithoutTopping );
        }
        return pizzas;
    }
}

public static void main( String[] args )
{
    String options[] = { "Pepperoni", "Bacon", "Cheese" };

    for( String onePizza : getPizzas( Arrays.asList( options ) ) )
    {
        System.out.println( onePizza );
    }
}

如何在Java中创建系统列表？

2 个答案: