Java中的简单数字计数器方法

Question

I am trying to create a method that will count the number of occurrences of digits in a string and record them into an array.

For example if the string entered into the method is "1223000", then counter[1] =1, counter[2] =2, counter[3] = 1, counter[0] = 3.

I keep getting a arrayindexoutofbounds error, here is the code I have soo far:

//method: count number of occurences for digits
    public static int[] count(String s){

        int[] counter = new int[10];

        for(int j= 0; j < s.length(); j++){
            if (Character.isDigit(s.charAt(j)))
                counter[s.charAt(j)] += 1;
        }

        return counter;
    }

Answer 1

See my comment for how to correct the issue.

You should also consider looping the characters of the string directly, rather than tracking the position in the string and using charAt .

For example,

public static int[] countDigits(final String str) {
  final int[] freq = new int[10];
  for (final char c : str.toCharArray()) {
    if (Character.isDigit(c)) {
      ++freq[c - '0'];
    }
  }
  return freq;
}

Testing the above using the below code results in no error ( java -ea DigitFreqTest ).

final String input = "1223000";
final int[] freq = countDigits(input);
assert freq[0] == 3 && freq[1] == 1 && freq[2] == 2 && freq[3] == 1;

---

Note the above does not support Unicode... in that case, you may wish to instead use Character.getNumericValue .

public static Map<Integer, Integer> countNumerals(final String str) {
  final Map<Integer, Integer> freq = new HashMap<Integer, Integer>(10);
  for (final char c : str.toCharArray()) {
    if (Character.isDigit(c)) {
      final int num = Character.getNumericValue(c);
      Integer occ = freq.get(num);
      if (occ == null) {
        occ = 0;
      }
      freq.put(num, occ + 1);
    }
  }
  return freq;      
}

Note I had to improvise using a Map<Integer, Integer> because Java does not inherently provide a multiset collection :-(

Answer 2

s.charAt(j) will give you the character number for that number, not it's integer value.

It's terribly corrected code, but you'll get the right idea: String s = "1223000"; int[] counter = new int[10]; String s = "1223000"; int[] counter = new int[10];

    for(int j= 0; j < s.length(); j++){

        if (Character.isDigit(s.charAt(j))) {
            int i = Integer.parseInt(s.substring(j, j+1));
            counter[i] += 1;
        }
            //unter[s.charAt(j)] += 1;
    }</code>

Answer 3

你得到ArrayIndexOutOfBoundsException ，因为s.charAt(j)在counter[s.charAt(j)]会返回数量char例如'1' ，然后转换字符使用ASCII所以为int char '1'将int 49等出数组索引。

Answer 4

计数器数组（int []计数器）的大小应为s的长度：int [] counter = new int [s.length（）];

Answer 5

If you guys want a counter for anything, just turn it into a string, then use the string method .length() to figure out the length, if that's what you're looking for. And if I'm not mistaken, the length() method returns an integer, so there you go. And as shown somewhere on this page, arrays have the same attribute, but that's for how many spaces there are that have things stored in them.