The problem of my method-convert each number type StringBuffer into an array of int.

When attempting to convert each symbol “9 8 7 6 5 4 3 2 1” a retracted space I get the result: {57,56,55,54,53,52,51,50,49}

How can I put into an array of the same number as in the line?

public static void main (String [] args) {
StringBuffer strB = new StringBuffer ( "9 8 7 6 5 4 3 2 1");
int arrInt = strBToArrInt (strB);
for (int i = 0; i & lt; arrInt.length; i ++) {System.out.print (arrInt [i] + "");}
} // end main
int [] strBToArrInt (StringBuffer strBuff) {
strBuff = new StringBuffer (. strBuff.toString () replace ( "", ""));
int [] arrInt = new int [strBuff.length ()];
for (int i = 0; i & lt; arrInt.length; i ++) {
      arrInt [i] = Integer.valueOf (strBuff.charAt (0)); // I think -problemma encoded.
  strBuff.deleteCharAt (0); // remove the null character
}
  return arrInt;
} // end strBToArrInt

Answer 1, Authority 100%

In your case, it returns the character ASCI-code. You should use the method digit . Replace

arrInt [i] = Integer.valueOf (strBuff.charAt (0));

In

arrInt [i] = Character.digit (strBuff.charAt (0), 10);

But this method is fit only for numbers. If you have numbers, you first need to break up the original string on an array of strings using the method split .

.

String [] stringsArray = strBuff.toString () trim () split ( "\\ s +").;

And then go and parse each line as the number of

int [] arrInt = new int [stringsArray.length ()];
for (int i = 0; i & lt; stringsArray.length; i ++) {
  arrInt [i] = Integer.parseInt (stringsArray [i]);
}

Answer 2, Authority 20%

Convert string into an array of characters:

char [] = String.toCharArray ();

Use the int [i] = Character.getNumericValue (char [i]); .
Of course, in your case, use the charAt (); , where in the same row can be converted to the int

.