将硬币数量转换为25分、10分、5分、1分的递归函数

Question

我有一个小项目，将硬币的数量转换成各自的25分，10分，5分，1分的数量。基本思想是这样的:

public static void coinCombo() {
int amount = getInt();
int quarter_cnt = amount/QUARTER;
int r1 = amount%QUARTER;
int dime_cnt = r1/DIME;
int r2 = r1%DIME;
int nickel_cnt = r2/NICKEL;
int r3 = r2%NICKEL;
System.out.println("Quarters: " + quarter_cnt);
System.out.println("Dimes: " + dime_cnt);
System.out.println("Nickels: " + nickel_cnt);
System.out.println("Pennies: " + r3);
}

然而，这在我看来相当笨重。我想知道如何递归地做到这一点。我看到我用前一个操作的余数来得到下一个。我觉得这可以做得更简洁。有人能举个递归的例子吗?

Answer 1

我会这样做。

static final int[] den = {25,10,5,1};
static final String[] names = {"Quarters", "Dimes","Nickels", "Pennies"};
public static void coinCombo(int amount) {
for (int i = 0; i < den.length; i++) {
System.out.printf("%s: %d%n", names[i], amount/den[i]);
amount %= den[i];
}
}
coinCombo(119);

打印

Quarters: 4
Dimes: 1
Nickels: 1
Pennies: 4

请注意，上面可以很容易地修改为包括更大面额的($10,$5和$1)。

但这是你的递归解。它还需要一个参数

coinCombo(232,0);
public static void coinCombo(int amount, int i) {
if (amount > 0) {
System.out.printf("%s: %d%n", names[i], amount/den[i]); 
coinCombo(amount % den[i], i+1);
}
}