这是作业:
编写一个名为
printStrings
的方法,它接受一个String和一个 作为参数的重复次数和打印给定的字符串 每次后有空格的次数。printStrings("abc", 5);
将打印以下输出:
abc abc abc abc abc
这就是我的尝试:
public class Apples {
public static String printStrings(String a) {
return (int i = 1; i <= 5; i++) {
System.out.print(a);
}
public static void main(String[] args) {
System.out.println(printStrings("abc"));
}
}
}
真的不知道该怎么做。
答案 0 :(得分:0)
稍微修改您的代码
public static void main(String[] args) {
System.out.println(printStrings("abc", 5));
}
public static String printStrings(String a, int numOfTimes) {
StringBuilder sb = new StringBuilder(""); // StringBuilder is better than concatenating Strings
for (int i = 1; i <= numOfTimes; i++) {
// System.out.println(sb);
sb.append(a); // append String
sb.append(" "); // append space
}
sb.replace(sb.length()-1, sb.length(), ""); // replace last space
System.out.println(sb.length());
return sb.toString(); // return String representation of StringBuilder sb
}
O / P:
19
abc abc abc abc abc
答案 1 :(得分:0)
public void printStrings(String a, int b) {
String printString = "";
for (int i = 0; i<b; i++){
printString = printString+" "+a;
}
System.out.println(printString);
}
答案 2 :(得分:0)
这样可行,但最后不会留出空格:
private static void printStrings(String str, int num) {
for (int i = 0; i < num; i++) {
System.out.print(str + (i == num - 1 ? "" : " "));
}
}
答案 3 :(得分:0)
public static String printStrings(String a, int count) {
String result = "";
for (int i = 0 ; i < count ; i++)
{
result = result + a;
if (i != (count-1))
result = result + " ";
}
return result;
}
public static void main(String[] args) {
System.out.println(printStrings("abc", 5));
}
答案 4 :(得分:-1)
public static String printStrings(String a,int b) {
String s = "";
for(int i = 1; i <= b; i++) {
s=s+a+" ";
}
return s;
}
public static void main(String[] args) {
System.out.println(printStrings("abc",5));
}