我正在尝试让三个数组同时运行一个循环。数组的长度不同，循环在最短的数组完成后退出。非常感谢任何帮助：
public class mortgagecalculator4 {
/**
* @param args the command line arguments
*/
public static void main(String[] args) {
// TODO code application logic here
double Loan1[] = {200000, 360, ((5.75 / 12) / 100)};//declares array
//and defines variables for first loan
double Loan2[] = {200000, 180, ((5.50 / 12) / 100)};//declares array
//and defines variables for second loan
double Loan3[] = {200000, 84, ((5.35 / 12) / 100)};//declares array
//and defines variables for third loan
double Payment1;//establishes variable for first payment
double Payment2;//establishes variable for second payment
double Payment3;//establishes variable for third payment
double Interest1;//establishes variable for interest for loan 1
double Interest2;//establishes variable for interest for loan 2
double Interest3;//establishes variable for interest for loan 3
double Balance1 = Loan1[0];//declares balance for loan 1
double Balance2 = Loan2[0];//declares balance for loan 2
double Balance3 = Loan3[0];//declares balance for loan 3
double Principal1;// declares variable for principal 1
double Principal2;// declares variable for principal 2
double Principal3;// declares variable for principal 3
Payment1 = (Loan1[0] * (Loan1[2])) / (1 - Math.pow(1 / (1 + Loan1[2]),
Loan1[1]));//calculates monthly payment for first loan
Payment2 = (Loan2[0] * (Loan2[2])) / (1 - Math.pow(1 / (1 + Loan2[2]),
Loan2[1]));//calculates monthly payment for second loan
Payment3 = (Loan3[0] * (Loan3[2])) / (1 - Math.pow(1 / (1 + Loan3[2]),
Loan3[1]));//calculates monthly payment for third lona
System.out.printf("The monthly payments are: Loan One:$%.2f ",
Payment1);//prints monthly payment amount for first loan
System.out.printf("Loan Two:$%.2f ",
Payment2);//prints monthly payment amount for second loan
System.out.printf("Loan Three:$%.2f\n",
Payment3);//prints monthly payment amount for third loan
System.out.printf("Each monthly payment breaks down as follows:\n");
//displays 'each monthly payment...'
for (int PaymentNumber = 1; PaymentNumber <= Loan1[1]
&& PaymentNumber <= Loan2[1] && PaymentNumber <= Loan3[1];
PaymentNumber++) {
//creates 'for' loop that continues until the LoanTerm variable
//is 0
Interest1 = Balance1 * Loan1[2];//Determines amt paid toward interest
Principal1 = Payment1 - Interest1;//Determines amt paid toward principal
Balance1 = Balance1 - Principal1;//Creates new balance for loop
PrintStream printf = System.out.printf(" Principal is: $%.2f", Principal1);
//prints out the amount of principal paid
System.out.printf(" Interest is: $%.2f", Interest1);
//prints otu the amount of interest paid
System.out.printf(" New balance is: $%.2f", Balance1);
//prints out the new balance amount
Interest2 = Balance2 * Loan2[2];//Determines amt paid toward interest
Principal2 = Payment2 - Interest2;//Determines amt paid toward principal
Balance2 = Balance2 - Principal2;//Creates new balance for loop
System.out.printf(" Principal is: $%.2f", Principal2);
//prints out the amount of principal paid
System.out.printf(" Interest is: $%.2f", Interest2);
//prints otu the amount of interest paid
System.out.printf(" New balance is: $%.2f", Balance2);
//prints out the new balance amount
Interest3 = Balance3 * Loan3[2];//Determines amt paid toward interest
Principal3 = Payment3 - Interest3;//Determines amt paid toward principal
Balance3 = Balance3 - Principal3;//Creates new balance for loop
System.out.printf(" Principal paid is: $%.2f", Principal3);
//prints out the amount of principal paid
System.out.printf(" Interest paid is: $%.2f", Interest3);
//prints otu the amount of interest paid
System.out.printf(" New balance is: $%.2f\n", Balance3);
//prints out the new balance amount
try {
Thread.sleep(500);//tells program to sleep for two seconds
} catch (InterruptedException ex) //catches any exceptions that are thrown
{
}
}
}
}
答案 0 :(得分：1)
为什么不
int minLength = Math.min(Math.min(arr1.size(), arr2.size()), arr3.size());
for (int i = 0; i < minLength; i++)
{
...
答案 1 :(得分：0)
好的，因为这显然是家庭作业，我将在这里提供指导。您遇到困难的情况是您有3个数组，需要在循环的同一次迭代中循环遍历每个数组。如果这不正确，请发表评论，我将更正这些陈述。所以这里的一般问题是你有一个循环需要处理集合中的X元素。这里要记住的是我们知道必须发生的迭代次数，这相当于最大收集长度。因此，如果集合中有10个元素，则循环需要持续到10个。
//一些伪代码
for（int i = 0; i＆lt; 10; i ++）// 10 ==最大集合尺寸
{...}
现在问题是你有x-1较小尺寸的集合，这会阻止你访问大于x sub i大小属性的元素。因此，为了解决这个问题，我们需要跟踪最后处理的元素索引。话虽这么说，首先要创建一个具有以下属性的循环：
for（int i = 0; i＆lt; collection.minSize; i ++）{...}
记录最后处理的索引，这将允许您提供后续循环以处理剩余的集合。
答案 2 :(得分：0)
当PaymentNumber为84时，您的当前循环将停止，因为这是满足循环条件的最高值（可以读作PaymentNumber <= 360 and PaymentNumber <= 180 and PaymentNumber <= 84，请注意PaymentNumber必须低于private static double handleLoan(double payment, double initialBalance, double[] loan) {
double balance = initialBalance;
for (int paymentNumber = 1; paymentNumber <= loan[1]; paymentNumber ++) {
interest = balance * loan[2];
double principal = payment - interest;
balance = balance - principal;
System.out.printf(" Principal is: $%.2f", principal);
System.out.printf(" Interest is: $%.2f", interest );
System.out.printf(" New balance is: $%.2f", balance );
}
return balance;
}
所有这些）。
似乎三个数组彼此无关，所以你可能只为每个数组使用3个不同的循环或更好，编写一个方法，将贷款（数组）作为参数并包含循环：
Balance1 = handleLoan(Payment1, Balance1, Loan1 );
然后像这样调用它（注意我使用你的变量，见底部的旁注）：
{{1}}
几个旁注：