此操作的目的是让用户每行输入一个数字,当用户不再希望继续时,他们应该能够输入一个空行,并且当发生这种情况时,程序应该给您一个最大的消息数字。
问题是我无法用空行使循环中断。我不确定该怎么做。我已经检查了其他问题以寻求解决方案,但找不到任何有帮助的方法。我也无法分配scan.hasNextInt() == null
....
我敢肯定有一个我没有想到的快速且合乎逻辑的解决方案。
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
System.out.println(Enter a number and press [Enter] per line, when you no longer wish to continue press [Enter] with no input.(empty line));
int x = 0;
while(scan.hasNextInt()){
int n = scan.nextInt();
if (n > x){
x = n;
}
}
System.out.println("Largets number entered: " + x);
}
}
答案 0 :(得分:2)
这应该可以解决您的问题:
import java.util.Scanner;
public class StackOverflow {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
System.out.println("Enter a number and press [Enter] per line, when you no longer wish to continue press [Enter] with no input.(empty line)");
int x = 0;
try {
while(!scan.nextLine().isEmpty()){
int num = Integer.parseInt(scan.nextLine());
if(num > x) {
x = num;
}
}
} catch (NumberFormatException e) {
e.printStackTrace();
}
System.out.println("Largest number entered: " + x);
scan.close();
}
}
答案 1 :(得分:2)
import java.util.*;
public class main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
System.out.println("Enter a number and press [Enter] per line, when you no longer wish to continue press [Enter] with no input.");
String str = scanner.nextLine();
int x = 0;
try {
while(!str.isEmpty()){
int number = Integer.parseInt(str);
if (number > x){
x = number;
}
str = scanner.nextLine();
}
}
catch (NumberFormatException e) {
System.out.println("There was an exception. You entered a data type other than Integer");
}
System.out.println("Largets number entered: " + x);
}
}