3 回答
TA贡献1951条经验 获得超3个赞
您可以简化程序的逻辑并编写如下内容,
public static void main(String[] args) {
Scanner keyboard = new Scanner(System.in);
List<String> nameList = new ArrayList<String>();
List<Double> scoreList = new ArrayList<Double>();
while (true) {
System.out.printf("Enter first name of student or done to finish: ");
String fname = keyboard.next();
if (fname.equals("done")) {
break;
}
System.out.printf("Enter last name of student: ");
String lname = keyboard.next();
nameList.add(fname + " " + lname);
System.out.println("Enter score: ");
scoreList.add(keyboard.nextDouble());
}
keyboard.close();
System.out.println("Names: " + nameList);
System.out.println("scores: " + scoreList);
}
TA贡献1816条经验 获得超4个赞
public static void main(String [] args)
{
Scanner keyboard = new Scanner(System.in);
double score;
boolean loopNaming=true;
int i=0;
ArrayList<String> name = new ArrayList<>();
while(loopNaming==true)
{
System.out.printf("Enter name of student or done to finish: ");
String input = keyboard.next();
if(input.equals("done"))
{
loopNaming = false;
}
else
{ name.add(input);
System.out.println("Enter score: ");
score = keyboard.nextDouble();
}
i=i+1; //no need to use
}
System.out.println(name);
}
您应该使用 adynamic list因为您不能在 Java 中调整数组的大小。第二点当用户给出“完成”时,你不应该把它放在列表中,所以在插入之前检查它。
TA贡献1824条经验 获得超6个赞
您声明了大小为 0 的 String 数组。这就是您不能向其中添加元素的原因。
import java.util.Scanner;
public class NameArray {
public static void main(String [] args){
Scanner keyboard = new Scanner(System.in);
double score[] = new double[10];
boolean loopNaming=true;
int i=0;
String namae;
String[] name = new String[10];
int count = 0;
while(loopNaming==true){
System.out.printf("Enter name of student or done to finish: ");
name[i] = keyboard.next();
if(name[i].equals("done")){
loopNaming = false;
}
else{
System.out.println("Enter score: ");
score[i] = keyboard.nextDouble();
count++;
}
i=i+1;
}
for(int j = 0; j < count; j++) {
System.out.println(name[j]+" "+score[j]);
}
}
}
试试这个代码,或者你可以选择任何其他数据结构。
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