在查找“n”个用户输入的平均值时尝试查找最小值和最大值时出错答案

【问题标题】：Error with trying to find Min&max when also finding average of "n" number of user inputs在查找“n”个用户输入的平均值时尝试查找最小值和最大值时出错
【发布时间】：2017-08-17 00:55:18
【问题描述】：

package HW2_Min_Max;

import java.util.Scanner; 

public class HW2_Min_Max {

    public static void main(String[] args) {

        Scanner myScanner = new Scanner(System.in);

        System.out.println("Please input a positive interger that indicates number of positive intergers ");

        int number = myScanner.nextInt();

        while (number <= 0) {
            System.out.println("Please input interger");
            number = myScanner.nextInt();
        }

        int i=1; //i is to store current iteration
        int sum=0; //sum is to store sum of the input 
        int x; //x is to store the user input

        while (i <= number){
            System.out.println("Please input a positive interger ");
            x = myScanner.nextInt();
            sum = sum + x;
            i++;
        }
        int average = sum/number;

        System.out.println("The average is " + average);

        int max = Integer.MIN_VALUE;
        int min = Integer.MAX_VALUE;

        if (number < min){
            min = number;
        }

        if (number > max) {
            max = number;
        }
        System.out.println("The minimum value is " + min);
        System.out.print( "and the maximum value is" + max);
        }
    }
}

1.^ 这是我遇到问题的地方，在 Netbeans 的最后一个大括号上，我收到一个错误，上面写着“预期的类、接口或枚举”，但我不知道为什么。请原谅我的无知，因为我是 java 的新手，更不用说编程了。

【问题讨论】：

看起来最后一个大括号实际上根本不需要。请记住{ 的数量应该等于} 的数量，您要确保打开的每个大括号最终都关闭。
正确、一致的缩进是你的朋友。
好的，这样就解决了让我运行代码的问题，但是我的实际输出有问题？我输入了5个正整数，平均值总是正确的，但是max和min的输出错误，我不知道为什么？
不要在 cmets 中重新定义您的问题。如果您还有其他问题，请单独提出。并包含更多细节。
不是直接针对您的问题，但“整数”这个词是正确的“整数”，如java.lang.Integer。

标签： java loops average minmax

【解决方案1】：

对于第一部分，我建议您使用do-while 循环来测试数字的数量。喜欢，

int number;
do {
    System.out.println("Please input a positive integer that "
            + "indicates number of positive integers ");
    number = myScanner.nextInt();
} while (number <= 0);

然后您需要在循环中设置min 和max （或存储您读取的所有值）。我更喜欢Math.max 和Math.min。我也会从0 算起。喜欢，

int i = 0; // i is to store current iteration
int sum = 0; // sum is to store sum of the input
int x; // x is to store the user input
int max = Integer.MIN_VALUE; // store the max user input
int min = Integer.MAX_VALUE; // store the min user input
while (i < number) {
    System.out.println("Please input a positive integer ");
    x = myScanner.nextInt();
    if (x > 0) { // make sure it's a positive integer
        min = Math.min(min, x);
        max = Math.max(max, x);
        sum += x;
        i++;
    }
}
int average = sum / number;
System.out.println("The average is " + average);
System.out.println("The minimum value is " + min);
System.out.println("and the maximum value is " + max);

【讨论】：

【解决方案2】：

它现在工作正常。我对它做了一些改变。请考虑一下。我希望它会有所帮助。

 Scanner myScanner = new Scanner(System.in);

    System.out.println("Please input a positive interger that indicates number of positive intergers ");


     int number = myScanner.nextInt();

    while (number <= 0) {
        System.out.println("Please input interger");
        number = myScanner.nextInt();
    }
    int[] arr = new int[number]; // Store values in array
    int i=0; //i is to store current iteration
    int sum=0; //sum is to store sum of the input
    int x; //x is to store the user input
    int max = 0;
    int min = 0;
       for(i=0;i<number;i++){
        System.out.println("Please input a positive interger ");
        arr[i] = myScanner.nextInt();
        sum = sum + arr[i];

    }
    int average = sum/number;

    System.out.println("The average is " + average);

    // Put initial values in min and max for comparing
    min =arr[0];
     max = arr[0];
for(i=1;i<number;i++)
{
    // Compare if values is less than arr[i]
    if (arr[i] < min){
        min = arr[i];
    }
// Compare if values is greater than arr[i]
    if (arr[i] > max) {
        max = arr[i];
    }
}
    System.out.print("The minimum value is " + min);
    System.out.println( " and the maximum value is " + max);

}
}

这是输出：

【讨论】：