计算字符串中字符的出现

Question

I tried to count the occurrence of alphabets in a string, but I substitue them with numbers to make it clearer. Then when I run that code, it doesnt display the results I want.I dont really know why...Please help!! Thank you so much!!

    Scanner Scanner1 = new Scanner(System.in);
    out.println("Please type in a string below.");
    String UserInput = Scanner1.nextLine();
    String Index = "12345";
    int length = 2;//Modified
    int[] count = new int[length];
    int length2 = 5; //Modified
    int n1 = 0;
    int n2 = 0;
    out.println(UserInput.charAt(n1));//Modified
    out.println(Index.charAt(n2));//Modified
    for (int i = 0; i < length; i++) {
        if (UserInput.charAt(n1) == Index.charAt(n2)) {
            n1++;
            count[length - (length - n1)]++;

        } else {
            n2++;
            if(n2==length2)
            {
                n2 = n2-length2;
            }
        }
    }

Answer 1

A relatively short and neat way to count a specific character in a string is using the return value of the replaceAll method:

public static int countChar(final String str, final char c) {
    return str.replaceAll("[^" + c + "]","").length();
}

The pattern [^x] ( x can be replaced with any char (or amount of different chars)) will match everything in a given String except x . So [^T] of TEST would replace E and S with the given replacement (which is "" (nothing)) and keeps the T s. The method would return TT . If you count that length, you'll receive the count of the searched character of the given string.

The example

System.out.println(countChar("TEST", 'T'));
System.out.println(countChar("TEST", 'E'));
System.out.println(countChar("TEST", 'S'));

prints

2
1
1

(keep in mind that is method is case sensitive)

Answer 2

use collections like Hashmap. Here Character stores every unique character encountered and Integer stores the count of every character whenever it is encountered.