The ASCII Problem

Question

我有一个“算法”，它的输入参数为整数k。例如，我将k = 54321作为参数传递。我想用这种方法：

将此整数转换为char数组，因此它将是[5,4,3,2,1]而不是54321，依此类推
按升序排列此数组，将每个字符添加到ArrayList
返回ArrayList

但是当涉及到for循环时，如果我通过char [position]获得项目，那么它会给我随机值。实际上，

public class Main {

    public static void main(String[] args) {
        System.out.println(sortMet(54321));
    }

    static ArrayList<Integer> sortMet(int k) {
        ArrayList<Integer> tab_cyfr = new ArrayList<>();
        char[] chars = String.valueOf(k).toCharArray();
        int n = chars.length;
        Arrays.sort(chars);

        for (int i = 0; i < n; i++) {
            tab_cyfr.add(chars[i]);
        }

        return tab_cyfr;
    }

}

这应该返回[1,2,3,4,5]，但返回[49，50，51，52，53]，这些数字来自哪里？

对变量命名感到抱歉，我正在对局部变量进行随机练习。

Answer 1

String.valueOf(k).toCharArray() returns the characters in the String as their char representation. As per ASCII table 49 is the char value of 1, 50 of 2, and so on.

You should convert k to digits, not to chars. This can be done by using % operation:

int k = 8421753;
ArrayList<Integer> digits = new ArrayList<>();
while (k > 0) {
    digits.add(k % 10);
    k /= 10;
}
Collections.sort(digits);
System.out.println(digits); // [1, 2, 3, 4, 5, 7, 8]

Answer 2

The ASCII Problem

It gets converted to a string, so the numbers are stored as ASCII values in the char array.

The problem is this line:

char[] chars = String.valueOf(k).toCharArray();

If you were to look up the ASCII values of the digits 1,2,3,4,5, you would see they are equal to 49, 50, 51, 52, 53,respectively.

How to Solve the Issue

To fix this issue, run a loop through the int number, take it apart digit by digit, and store them as ints, not chars, and then just sort your int[].

Answer 3

Replacing your ArrayList<Integer> to ArrayList<Character> solve this problem because it store characters '1', '2', etc not their Integer values which are 49, 50...

So below code return what you expect: [1, 2, 3, 4, 5]

public static void main(String[] args) {
    System.out.println(sortMet(54321));
}

static ArrayList<Character> sortMet(int k) {
    ArrayList<Character> tab_cyfr = new ArrayList<>();
    char[] chars = String.valueOf(k).toCharArray();
    int n = chars.length;
    Arrays.sort(chars);

    for (int i = 0; i < n; i++) {
        tab_cyfr.add(chars[i]);
    }

    return tab_cyfr;
}

Answer 4

这些数字是数组中字符的ASCII码。例如，字符0为48，字符1为49，依此类推。似乎您正在使用Integer的数组列表，因此list元素将像标准int和long整数一样打印。要解决此问题，请将ArrayList<Integer>替换为ArrayList<Character>。

Answer 5

此行无法编译：

tab_cyfr.add(chars[i]);

所以您得到[49，50，51，52，53]很奇怪。
即使编译，它也会添加chars[i]的ascii值，
但是您想要它的数值。
一种获得它的方法是：

tab_cyfr.add(Integer.parseInt("" + chars[i]));

Answer 6

toCharArray() method returns the character array and when you sort array is sorted by its ASCII value. So those random numbers are ASCII values.
To convert char into Integer:-

static ArrayList<Integer> sortMet(int k) {
        ArrayList<Integer> tab_cyfr = new ArrayList<>();
        char[] chars = String.valueOf(k).toCharArray();
        int n = chars.length;


        for (int i = 0; i < n; i++) {
            tab_cyfr.add(Character.getNumericValue(chars[i]));
        }
        Collections.sort(tab_cyfr);
        return tab_cyfr;
    }

Answer 7

This code should work instead.

public static void main(String[] args) {
    System.out.println(sortMet(54321));
}

static ArrayList<Integer> sortMet(int k) {
    ArrayList<Integer> tab_cyfr = new ArrayList<>();
    char[] chars = String.valueOf(k).toCharArray();
    int n = chars.length;
    Arrays.sort(chars);

    for (int i = 0; i < n; i++) {
        tab_cyfr.add(Character.getNumericValue(chars[i]));
    }
    return tab_cyfr;
}

Character.getNumericValue(chars[i]) ensures that you push the actual char value into tab_cyfr and not the ASCII value of it.

char [i]返回错误的数字

7 个答案:

The ASCII Problem

How to Solve the Issue