如何在Java中创建一个系统列表

Question

嘿，伙计们，我试图在Java中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:假设你总共有6种配料来做披萨。你能做出多少种不同的5种配料的披萨?

例如:

[意大利香肠，培根，菠萝，洋葱，蘑菇，辣椒][意大利辣肠，培根，菠萝，洋葱，蘑菇，奶酪]等等…

在开发过程中，我一直在练习3个可能的选项，而不是6个，因为我知道将有三个总组合，每个组合有两个元素，使我的多维数组在这个时候很容易生成。

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列表来存储这些值。

int[# possible combination][# possible values]

我当前的代码假设我们已经知道可能的组合(尽管我们当然不知道)，并且无论如何我们总是可以确定值(可以制作多少个5、4、3、2个披萨)

我如何能够从options[]中选择一个元素，同时确保没有其他数组已经包含你试图插入的那些元素。我会尝试数组。包含或数组。但是我不知道插入什么来比较，combos[pizzas-1] ?

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

Answer 1

一种方法是只看你还没有考虑过的配料。举个例子，当你从[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，你可能会决定跳过奶酪，然后选择辣椒，然后跳过蘑菇和洋葱，选择菠萝和培根。当然，你需要留下足够的配料来完成你的披萨，所以你不能跳过奶酪、辣椒、蘑菇、洋葱和菠萝。示例代码在JavaScript中使用递归算法给你一个概念。

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。有两列是配料。如果我想要一个纯奶酪披萨呢?

这是一个使用递归构建不同数量配料的披萨的解决方案。

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 );
    }
}