我正在为我的Java类编写一个程序，要求这样做：
编写一个降雨类，将12个月中的每个月的总降雨量存储为两倍的数组。该程序应具有返回以下内容的方法：
年总降雨量
月平均降雨量
下雨最多的月份
下雨最少的月份
在一个完整的程序中演示该类。 （不接受月降雨量的负数）
import java.util.Scanner;
import java.io.*;
public class apples{
public static void main (String[] args){
Scanner kenny = new Scanner(System.in);
double rain[]=new double[13];
double sum = 0;
double avg =0;
double most =0;
double least =0;
System.out.println("Your local weather man here getting paid to tell you the wrong weather!!");
System.out.println("");
System.out.println("Please enter in the following rainfall for the months ahead: ");
System.out.println("Month\tRainfall (In inches)");
System.out.print("January: ");
rain [0] = kenny.nextDouble();
System.out.print("February: ");
rain [1] = kenny.nextDouble();
System.out.print("March: ");
rain [2] = kenny.nextDouble();
System.out.print("April: ");
rain [4] = kenny.nextDouble();
System.out.print("May: ");
rain [5] = kenny.nextDouble();
System.out.print("June: ");
rain [6] = kenny.nextDouble();
System.out.print("July: ");
rain [7] = kenny.nextDouble();
System.out.print("August: ");
rain [8] = kenny.nextDouble();
System.out.print("September: ");
rain [9] = kenny.nextDouble();
System.out.print("October: ");
rain [10] = kenny.nextDouble();
System.out.print("November: ");
rain [11] = kenny.nextDouble();
System.out.print("December: ");
rain [12] = kenny.nextDouble();
//(Or rain[] = 1,2,3,4,5,6,7,8,9,10,11,12);
sum = rain[0] + rain[1] + rain[2] + rain[3] + rain[4] + rain[5] + rain[6] + rain[6] + rain[7] + rain[8] + rain[9] + rain[10] + rain[11] + rain[12] ;
avg = (rain[0] + rain[1] + rain[2] + rain[3] + rain[4] + rain[5] + rain[6] + rain[6] + rain[7] + rain[8] + rain[9] + rain[10] + rain[11] + rain[12]) / 12;
System.out.println("The sum of all the rain is: " + sum);
System.out.println("The average rainfall was:" + avg + " inches");
System.out.print("The month with the most rain was: ");
}
private static void getMaxValue(double[] rain) {
getMaxValue(rain);
System.out.println(getMaxValue(rain));
System.out.println("The month with the least rain was: ");
}
private static void getMinValue(double[] rain) {
getMinValue(rain);
System.out.println(getMaxValue(rain));
}}
我已经准备好大部分。我只是想知道如何从输入的数字中获取“最大”和“最小”。
任何帮助将是巨大的！
最佳答案
您可以通过循环数组找到max或min。并将返回类型void更改为double，这样方法将返回max rain；
private static double getMaxValue(double[] rain) {
double max=0;
for(double i : rain){
if(i>max){
max=i;
}
}
return max;
}
并用作
System.out.println(getMaxValue(rain));
同样的分钟
private static double getMinValue(double[] rain) {
double min=Double.MAX_VALUE;
for(double i : rain){
if(i<min){
min=i;
}
}
return min;
}
但是在您的代码中有很多错误
1）
double rain[]=new double[13];
这应该是
double rain[]=new double[12];
因为这是数组长度，所以您有12个月的时间。
2）你错过了
rain [3]
3）您分配给13的索引应该是12。
rain [13] = kenny.nextDouble(); --> rain [12] = kenny.nextDouble();
因此，这是完整的示例。
public class apples {
public static void main(String[] args) {
Scanner kenny = new Scanner(System.in);
double rain[] = new double[12];
double sum = 0;
double avg = 0;
double most = 0;
double least = 0;
System.out.println("Your local weather man here getting paid to tell you the wrong weather!!");
System.out.println("");
System.out.println("Please enter in the following rainfall for the months ahead: ");
System.out.println("Month\tRainfall (In inches)");
System.out.print("January: ");
rain[0] = kenny.nextDouble();
System.out.print("February: ");
rain[1] = kenny.nextDouble();
System.out.print("March: ");
rain[2] = kenny.nextDouble();
System.out.print("April: ");
rain[3] = kenny.nextDouble();
System.out.print("May: ");
rain[4] = kenny.nextDouble();
System.out.print("June: ");
rain[5] = kenny.nextDouble();
System.out.print("July: ");
rain[6] = kenny.nextDouble();
System.out.print("August: ");
rain[7] = kenny.nextDouble();
System.out.print("September: ");
rain[8] = kenny.nextDouble();
System.out.print("October: ");
rain[9] = kenny.nextDouble();
System.out.print("November: ");
rain[10] = kenny.nextDouble();
System.out.print("December: ");
rain[11] = kenny.nextDouble();
//(Or rain[] = 1,2,3,4,5,6,7,8,9,10,11,12);
sum = rain[0] + rain[1] + rain[2] + rain[3] + rain[4] + rain[5] + rain[6] + rain[7] + rain[8] + rain[9] + rain[10] + rain[11];
avg = sum / 12;
System.out.println("The sum of all the rain is: " + sum);
System.out.println("The average rainfall was:" + avg + " inches");
most =getMaxValue(rain);
least=getMinValue(rain);
System.out.println("The max rain is: " + most);
System.out.println("The min rain is: " + least);
}
private static double getMaxValue(double[] rain) {
double max = 0;
for (double i : rain) {
if (i > max) {
max = i;
}
}
return max;
}
private static double getMinValue(double[] rain) {
double min = Double.MAX_VALUE;
for (double i : rain) {
System.out.println(i);
if (i < min) {
min = i;
}
}
System.out.println(min);
return min;
}
}
但是您可以使用包含所有月份的数组。这样做的好处是可以动态循环而不是硬接线。当输入为负数时，您可以发出警告。
public class apples {
public static void main(String[] args) {
Scanner kenny = new Scanner(System.in);
double rain[] = new double[12];
double sum = 0;
double avg = 0;
double most = 0;
double least = 0;
System.out.println("Your local weather man here getting paid to tell you the wrong weather!!");
System.out.println("");
System.out.println("Please enter in the following rainfall for the months ahead: ");
System.out.println("Month\tRainfall (In inches)");
String months[]={"January", "February", "March", "April", "May", "June", "July", "August", "September", "October", "November", "December"};
for (int i=0;i<months.length;i++) {
System.out.println(months[i]+" :");
double val = kenny.nextDouble();
while(val<0){
System.out.println("negatives not allowed ! enter again");
val = kenny.nextDouble();
}
rain[i]=val;
sum+=val;
}
avg = sum / 12;
System.out.println("The sum of all the rain is: " + sum);
System.out.println("The average rainfall was:" + avg + " inches");
most =getMaxValue(rain);
least=getMinValue(rain);
System.out.println("The max rain is: " + most);
System.out.println("The min rain is: " + least);
}
private static double getMaxValue(double[] rain) {
double max = 0;
for (double i : rain) {
if (i > max) {
max = i;
}
}
return max;
}
private static double getMinValue(double[] rain) {
double min = Double.MAX_VALUE;
for (double i : rain) {
System.out.println(i);
if (i < min) {
min = i;
}
}
System.out.println(min);
return min;
}
}