你好我正在研究一个项目并遇到了一个我无法弄清楚的问题。
我有3个数组都是double ... hours []，rate []，prevHrs []。我将小时和前提复制到一个名为monthhours []的新数组中，现在我需要重新排列它，以便将小时数组中的元素分别放入适当的位置。我需要prevhrs的前3个元素，然后是小时的第一个元素，然后是prevhrs的4,5,6元素和小时的第2个元素，依此类推。然后，在完成所有操作之后，我需要从这个新的monthHours数组中获取每个元素，并将其乘以rate数组中的元素。基本上费率[0]乘以mountHours [0] [1] [2] [3]然后费率[1]乘以monthHours [4] [5] [6] [7]
如果有人能提供帮助，我将不胜感激。这就是我到目前为止所做的事情，而且我已经问过全能的谷歌，但我们还没能找到我想要的东西。
String [] names = {"Barry", "Lindsey", "Arron", "Brent"};
double [] hours = {40.0, 37.5, 39.5, 18.5};
double [] rate = {15.67, 8.90, 8.90, 12.33};
double [] prevHrs = {32.0, 40.0, 39.0, 28.5, 31.5, 38.0, 40.0, 24.0, 36.0, 40.0, 40.0, 22.5};
double [] monthHours = new double[16];
double mostHours;
mostHours =FinalMethods.getHighest(hours);
double pay[] = FinalMethods.calculatePay(hours, rate);
double totalPay[] = FinalMethods.calculateMonthpay(monthHours, rate);
System.out.printf("The employee who Worked the most hours is %s" + " With %.2f" + "hours\n",names[0], mostHours);
System.out.println(" ");
System.out.println("Name \tHours \tRate \tTotal");
for(int i=0;i<names.length;i++)
System.out.printf(names[i] +"\t"+ hours[i] +"\t"+ rate[i] +"\t"+ pay[i]+"\n" );
System.arraycopy(hours, 0, monthHours, 0, hours.length);
System.arraycopy(prevHrs, 0, monthHours, hours.length, prevHrs.length);
System.out.println(" ");
System.out.println("Employee Monthly Pay");
System.out.println(" ");
System.out.println("Name \tTotal");
for(int i=0;i<names.length;i++)
System.out.printf(names[i] +"\t"+ totalPay[i]+"\n" );
}
// Calculate Pay
public static double[] calculatePay(double[] hours, double[] rate){
double[] pay = new double[hours.length];
for (int i = 0; i < hours.length; i++) {
pay[i] = hours[i] * rate[i];
}
return pay;
}
// Calculate Monthly Pay
public static double[] calculateMonthpay(double[] monthHours, double[] rate){
double[] totalPay = new double[monthHours.length];
for(int i = 0; i < monthHours.length; i++){
totalPay[i] = monthHours[i] * rate[i] % rate[rate.length];
}
return totalPay;
}
答案 0 :(得分：1)
如果你离开看似安全的阵列并进入面向对象编程的可怕深渊，你可以让你的生活更轻松。
在这里，我拿着一个Employees数组，并让他们计算自己的工资。
class Employee {
final String name;
final double hours;
final double rate;
final double[] prevHours;
public Employee(String name, double hours, double rate, double[] prevHours) {
this.name = name;
this.hours = hours;
this.rate = rate;
this.prevHours = prevHours;
}
public String getName() {
return name;
}
public double getHours() {
return hours;
}
public double getRate() {
return rate;
}
public double[] getPrevHours() {
return prevHours;
}
public double getPay() {
return rate * hours;
}
public double getMonthsPay() {
double monthsPay = getPay();
for (double h : prevHours) {
monthsPay += rate * h;
}
return monthsPay;
}
}
private Employee getMostHours(Employee[] employees) {
Employee most = null;
for (Employee employee : employees) {
if (most == null || employee.getHours() > most.getHours()) {
most = employee;
}
}
return most;
}
public void test() {
System.out.println("Hello");
Employee[] employees = {
new Employee("Barry", 40.0, 15.67, new double[]{32.0, 40.0, 39.0}),
new Employee("Lindsey", 37.5, 8.90, new double[]{28.5, 31.5, 38.0}),
new Employee("Arron", 39.5, 8.90, new double[]{28.5, 31.5, 38.0}),
new Employee("Brent", 18.5, 12.33, new double[]{28.5, 31.5, 38.0})};
Employee mostHours = getMostHours(employees);
System.out.printf("The employee who Worked the most hours is %s" + " With %.2f" + "hours\n", mostHours.getName(), mostHours.getHours());
System.out.println();
System.out.println("Name \tHours \tRate \tTotal \tMonth");
for (Employee employee : employees) {
System.out.println(employee.getName()
+ "\t" + employee.getHours()
+ "\t" + employee.getRate()
+ "\t" + employee.getPay()
+ "\t" + employee.getMonthsPay());
}
}
答案 1 :(得分：1)
我同意面向对象编程方法更好，但是因为人们可能会手动执行此操作，所以我为您的第一个用例编写了一个小例子。
所以这是通过添加3个元素，然后添加另外1个元素来组合数组。它不漂亮，但应该有希望说明这个想法：
public static void main(String[] args) {
double [] hours = {40.0, 37.5, 39.5, 18.5};
double [] prevHrs = {32.0, 40.0, 39.0, 28.5, 31.5, 38.0, 40.0, 24.0, 36.0, 40.0, 40.0, 22.5};
// 3 elements of prevHrs, 1 element of hours
int newSize = hours.length + prevHrs.length;
double[] combinedArray = new double[newSize];
int prevHrsOffset = 0;
for(int i= 0; i < hours.length; i++) {
for( int j=0; j<3; j++) { // 3 elements
combinedArray[i+prevHrsOffset] = prevHrs[prevHrsOffset];
prevHrsOffset ++;
// TODO insert safety check for arrayoutofbounds
}
combinedArray[i+prevHrsOffset] = hours[i];
}
for(int i=0; i < combinedArray.length; i++) {
System.out.println(combinedArray[i]);
}
}
输出结果为：
32.0, 40.0, 39.0, 40.0, 28.5, 31.5, 38.0, 37.5, 40.0, 24.0, 36.0, 39.5, 40.0, 40.0, 22.5, 18.5
请注意，这不是一种非常安全的方法，您必须防止索引违规等。
您可以使用列表添加（至少这种方式索引不是问题）。
此外，小时和上一个小时的长度可能不同，然后组合不再工作。
希望能回答你的问题并告诉你如何做到这一点。
以类似方式计算总薪酬：
double [] rate = {15.67, 8.90, 8.90, 12.33};
double[] totalPay = new double[newSize]; // the total pay based on that other array
int combinedArrayOffset = 0;
for (int i = 0; i < rate.length; i++) {
for (int j = 0; j < 4; j++) { // 4 elements
totalPay[combinedArrayOffset] = rate[i] * combinedArray[combinedArrayOffset];
combinedArrayOffset++;
// TODO insert safety check for arrayoutofbounds
}
}
System.out.println("total pay:");
for(int i=0; i < totalPay.length; i++) {
System.out.print(totalPay[i] + ", ");
}
答案 2 :(得分：0)
我强烈建议习惯于面向对象的编程，正如已经建议的那样，但是既然你在寻求当前代码的帮助，那么这就是我的解决方案 - 写在＆＃34;看似安全的数组基础＆＃34;：< / p>
<强>代码：
public class FinalMethods {
public static void main(String[] args) {
String[] names = {"Barry", "Lindsey", "Arron", "Brent"};
double[] hours = {40.0, 37.5, 39.5, 18.5};
double[] rate = {15.67, 8.90, 8.90, 12.33};
double[] prevHrs = {
32.0, 40.0, 39.0,
28.5, 31.5, 38.0,
40.0, 24.0, 36.0,
40.0, 40.0, 22.5
};
double[] monthHours;
{
// we have name.length employees, and for each 3 prevHrs entries and 1 hours entry
monthHours = new double[names.length * 4];
for (int i = 0; i < names.length; i++) {
// append/copy range of 3 prevHrs entries to monthHours
System.arraycopy(prevHrs, i * 3, monthHours, i * 4, 3);
// append/copy range of 1 hours entries to monthHours
System.arraycopy(hours, i, monthHours, i * 4 + 3, 1);
// equivalent to:
/*
monthHours[i * 4] = prevHrs[i * 3];
monthHours[i * 4 + 1] = prevHrs[i * 3 + 1];
monthHours[i * 4 + 2] = prevHrs[i * 3 + 2];
monthHours[i * 4 + 3] = hours[i];
*/
}
}
int mostHoursIndex = FinalMethods.getHighest(hours);
double pay[] = FinalMethods.calculatePay(hours, rate);
double totalPay[] = FinalMethods.calculateMonthPay(monthHours, rate);
// most worked
System.out.printf("The employee who worked most hours is %s with %.2f hours\n", names[mostHoursIndex], hours[mostHoursIndex]);
System.out.println();
// print pay table
System.out.println("\n-- Employee Pay --\n");
System.out.printf("%-10s %-10s %-10s %-10s\n", "Name", "Hours", "Rate", "Total");
System.out.println();
for (int i = 0; i < names.length; i++) {
System.out.printf("%-10s %-10.2f %-10.2f %-10.2f\n", names[i], hours[i], rate[i], pay[i]);
}
System.out.println();
// print monthly pay table
System.out.println("\n-- Employee Monthly Pay --\n");
System.out.printf("%-10s %-10s\n", "Name", "Total");
System.out.println();
for (int i = 0; i < names.length; i++) {
System.out.printf("%-10s %-10.2f\n", names[i], totalPay[i]);
}
}
public static int getHighest(double... values) {
// result will be -1 for empty arrays
int result = -1;
// EVERY value should be > max when starting, thus:
double max = Double.NEGATIVE_INFINITY;
for (int i = 0; i < values.length; i++) {
double value = values[i];
if (value > max) {
// found a higher entry, mark index and update max
max = value;
result = i;
}
}
// return index of highest entry
return result;
}
public static double[] calculatePay(double[] hours, double[] rate) {
double[] result = new double[rate.length];
// for each employee
for (int i = 0; i < result.length; i++) {
// sum his hours (1 entry) * his rate
result[i] = hours[i] * rate[i];
}
return result;
}
public static double[] calculateMonthPay(double[] monthHours, double[] rate) {
double[] result = new double[rate.length];
// for each employee
for (int i = 0; i < result.length; i++) {
// sum his monthHours (4 entries) * his rate
double monthPay = 0;
for (int j = 0; j < 4; j++) {
monthPay += monthHours[i * 4 + j] * rate[i];
}
result[i] = monthPay;
}
return result;
}
}
<强>输出：
The employee who worked most hours is Barry with 40,00 hours
-- Employee Pay --
Name Hours Rate Total
Barry 40,00 15,67 626,80
Lindsey 37,50 8,90 333,75
Arron 39,50 8,90 351,55
Brent 18,50 12,33 228,11
-- Employee Monthly Pay --
Name Total
Barry 2366,17
Lindsey 1205,95
Arron 1241,55
Brent 1491,93