java – 打印哪些硬币用于赚取给定金额

2019年7月22日 217次阅读

我试图使用递归来找到给定数量的最小金币数量.我的代码能够列出所需的最小硬币数量,但我似乎无法找到打印出哪些硬币用于提出解决方案的方法.我搜索并找到了类似的例子,但我似乎无法正确地应用它.

这是我到目前为止：

import java.util.*;

public class Coins{

    public static int findMinCoins(int[] currency, int amount) {
        int i, j, min, tempSolution;

        min = amount;

        for (i = 0; i < currency.length; i++) {
            if (currency[i] == amount) {
                return 1;
            }
        }

        for (j = 1; j <= (amount / 2); j++) {
            tempSolution = findMinCoins(currency, j) + findMinCoins(currency, amount - j);
            if (tempSolution < min) {
                min = tempSolution;
            }
        }
        return min;
    }

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int[] USA =
        {1, 5, 10, 25, 50};
        System.out.println("Please enter an integer amount.");
        int amount = in.nextInt();
        int minCoins = findMinCoins(USA, amount);
        System.out.println("The minimum number of coins to make " + amount + " in United States currency is " + minCoins + ".");
        System.out.println("The coins used were:");
        /*Print coins used to find minCoins.*/
        in.close();
    }
}

到目前为止运行的代码示例：

Please enter an integer amount.
17
The minimum number of coins to make 17 in United States currency is 4.
The coins used were:

如果有人能给我一些关于如何做到这一点的见解,我将不胜感激.

最佳答案 我认为这应该与您想要实现的目标完全一致.只需调用public static int findMinCoins(arg1,arg2),它将输出使用递归算法使用的最小数量的硬币和所有特定硬币(它将显示其出现次数).

public static int findMinCoins(int[] currency, int amount) {
    int min = findMinCoins(currency, amount, 0);
    System.out.print("The coins used were: ");
    return min;
}
private static int findMinCoins(int[] currency, int amount, int min){
    int number, value1, value2;
    int min1 = min;
    for(int i=currency.length-1; i>=0; i--) {
        if (amount>=currency[i]){
            amount = amount - currency[i];
            System.out.print(currency[i] + " ");
            min1 = findMinCoins(currency, amount, min1);
            return ++min1;
        }
    }
    return min1;
}
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