Java 如何对数组中的字符串进行排序?下降/上升 扫描仪输入=新扫描仪(System.in); String[]String_数组=新字符串[5]; for(int i=0;i
如何对数组中的字符串进行排序?下降/上升Java 如何对数组中的字符串进行排序?下降/上升 扫描仪输入=新扫描仪(System.in); String[]String_数组=新字符串[5]; for(int i=0;i,java,Java,如何对数组中的字符串进行排序?下降/上升 扫描仪输入=新扫描仪(System.in); String[]String_数组=新字符串[5]; for(int i=0;i
扫描仪输入=新扫描仪(System.in);
String[]String_数组=新字符串[5];
for(int i=0;i
扫描仪输入=新扫描仪(System.in);
String[]String_数组=新字符串[5];
for(int i=0;i
您会遇到什么错误?您的代码以什么方式不起作用?您似乎正在尝试对名为args
的数组进行排序,但您在哪里定义它?(提示:不在显示的代码中。)“args”是“public void main(String[]args)”中的默认名称,因此如果您是这个意思,则应该插入完整的main方法。
Scanner input = new Scanner(System.in);
String[] string_array = new String[5];
for (int i = 0; i < string_array.length; i++)
{
System.out.println("Please enter a string to sort:");
string_array[i] = input.next();
}
System.out.println ("\nSorting options:\nA.Ascending\nB.Descending\n\nPlease enter your choice in specific letter:");
Scanner s = new Scanner (System.in);
String a = s.next();
if (a.equals("a")){
Arrays.sort(args);
System.out.println ();
for(int i = 0; i < string_array.length; i++)
System.out.println(string_array[i]);
Scanner input = new Scanner(System.in);
String[] string_array = new String[5];
for (int i = 0; i < string_array.length; i++)
{
System.out.println("Please enter a string to sort:");
string_array[i] = input.next();
}
System.out.println ("\nSorting options:\nA.Ascending\nB.Descending\n\nPlease enter your choice in specific letter:");
Scanner s = new Scanner (System.in);
String a = s.next();
if (a.equalsIgnoreCase("a")){
//I think this should be passed into string_array
Arrays.sort(string_array);
}else if(a.equalsIgnoreCase("b")){
Arrays.sort(string_array, new Comparator<String>() {
public int compare(String o1, String o2) {
return o1.compareTo(o2);
}
});
}
System.out.println ();
for(int i = 0; i < string_array.length; i++)
System.out.println(string_array[i]);
}