题目

给出一个随机数(long)，您必须以相反的顺序在数组中返回此数字的数字。

例：

348597 => [7,9,5,8,4,3]

测试用例：

import org.junit.Test;
import static org.junit.Assert.assertArrayEquals;

public class DigitizeExampleTests {
  @Test
  public void tests() {
    assertArrayEquals(new int[] {1, 3, 2, 5, 3}, Kata.digitize(35231));
  }
}

解题

My

public class Kata {
  public static int[] digitize(long n) {
    char[] arrChar = String.valueOf(n).toCharArray();
    int[] arrInt = new int[arrChar.length];
    for (int i = 0; i < arrChar.length; i++) {
      arrInt[i] = Integer.parseInt(String.valueOf(arrChar[arrChar.length-i-1]));
    }
    return arrInt;
  }
}

Other

public class Kata {
  public static int[] digitize(long n) {
        return new StringBuilder().append(n)
                                  .reverse()
                                  .chars()
                                  .map(Character::getNumericValue)
                                  .toArray();
  }
}

import java.lang.Math;
public class Kata {
  public static int[] digitize(long n) {
    String s = String.valueOf(n);
    int length = s.length();
    int[] array = new int[length];
    for ( int i = 0; i < length; i++){
      array[i] = (int) (s.charAt(length - i - 1)) - 48;
    }
    return array;
  }
}

这方法也还不错。

后记

一拿到这题就知道，答案肯定不唯一，我的代码还可以优化，嘎嘎。

    public static int[] digitize(long n) {
        String m = String.valueOf(n);
        int[] arrInt = new int[m.length()];
        for (int i = 0; i < m.length(); i++) {
            arrInt[i] = m.charAt(m.length() - i - 1) - 48;
        }
        return arrInt;
    }