提问人:coolkid9090 提问时间:3/20/2021 更新时间:3/20/2021 访问量:176
Java 循环扫描器输入,直到 enter 或 “e” 不带 .split
Java loop scanner input until enter or "e" without .split
问:
public class test {
private static Scanner userpress = new Scanner(System.in);
public static void main(String[] args) {
int r = 0;
int h = 0;
System.out.println("---------------------------------");
System.out.println("write your two numbers (numerator, denominator)");
userpress.useDelimiter("\\s"); // here
while (userpress.hasNextInt()) {
r = userpress.nextInt();
h = userpress.nextInt();
// userpress.nextLine(); // remove
int x = r / h;
System.out.print(x + " ");
}
}
private static void message() {
System.out.println("user pressed e");
}
}
该程序要求用户至少输入 2 个输入,第一个输入是分子,第二个输入是分母,然后它将给出分子/分母的值并将其打印出来。
这是我希望能够做的:我希望能够写下尽可能多的数字,例如,输出应该是,如果我写,输出应该是。但是,这仅适用于偶数。例如,如果我写不偶数,那么程序就会崩溃,因为 2 没有被任何东西除以。我希望程序删除最后一个输入,如果它不均匀,例如,如果输入是 ,那么输出应该是 因为 10/5=2 并且最后一个输入应该被丢弃。我该如何解决这个问题?
问题 2:例如,如果我在数字后面写一个;,然后我希望输出是我不想使用 .split,我必须使用 和10 5 20 4
2, 5
10 5 20 4 30 5
2, 5, 6
10 5 3
10 5 3
2
e
10 5 20 4 e
2, 5 user pressed e
hasnext...
next()
答:
0赞
Anil Demir
3/20/2021
#1
您可以使用正则表达式来检查空格,而不是在空格之前取唯一的第一部分。 喜欢这个:
import java.util.Scanner;
public class Main {
private static Scanner userpress = new Scanner(System.in);
public static void main(String[] args) {
String first;
String second;
System.out.println("---------------------------------");
System.out.println("write your two numbers (numerator, denominator)");
userpress.useDelimiter("\\s"); // here
while (userpress.hasNextLine()) {
first = userpress.nextLine();
second = userpress.nextLine();
if(first.contains(" ") || second.contains(" ")) {
String[] firstParts = first.split(" ");
String[] secondParts = second.split(" ");
first = firstParts[0];
second = secondParts[0];
}
// userpress.nextLine(); // remove
int x = Integer.valueOf(first) / Integer.valueOf(second);
System.out.print(x + " ");
}
}
private static void message() {
System.out.println("user pressed e");
}
}
0赞
Arvind Kumar Avinash
3/20/2021
#2
代码的问题在于,您正在扫描两个整数,只需检查 。每次扫描整数时都应该进行此检查。userpress.hasNextInt()
工作代码:
import java.util.Scanner;
public class Main {
public static void main(String args[]) {
int r = 0;
int h = 0;
System.out.print("Write your two numbers (numerator, denominator): ");
Scanner userpress = new Scanner(System.in);
userpress.useDelimiter("\\s");
while (userpress.hasNextInt()) {
r = userpress.nextInt();
if (userpress.hasNextInt()) {
h = userpress.nextInt();
int x = r / h;
System.out.print(x + " ");
} else {
break;
}
}
String s = userpress.next();
if ("e".equalsIgnoreCase(s)) {
message();
}
}
private static void message() {
System.out.println("user pressed e");
}
}
示例运行:
Write your two numbers (numerator, denominator): 10 5 3 e
2 user pressed e
另一个示例运行:
Write your two numbers (numerator, denominator): 10 5 3
2
另一个示例运行:
Write your two numbers (numerator, denominator): 10 5 20 5
2 4
另一个示例运行:
Write your two numbers (numerator, denominator): 10 5 20 5 e
2 4 user pressed e
0赞
Giorgi Tsiklauri
3/20/2021
#3
或者,如果您不介意使用更多空间,您可以想出一个简单的算法,例如:
public class Main {
private static Scanner userpress = new Scanner(System.in);
public static void main(String[] args) {
System.out.println("---------------------------------");
System.out.println("write your two numbers (numerator, denominator)");
userpress.useDelimiter("\\s");
List<Integer> numbers = new ArrayList<>();
while(userpress.hasNextInt()) {
numbers.add(userpress.nextInt());
}
for (int i=0; i<numbers.size()-1; i+=2) {
int x = numbers.get(i)/numbers.get(i+1);
System.out.print(x + " ");
}
}
}
评论
int
e