使用Java的字符串中的字符的次数？

Question

I searched for method to calculate how many times a character is in String and found a good solution

temp.length() - temp.replaceAll("T", "").length()

here,we are calculating the number of time 'T' is in temp...

Problem is,it is not working properly for '.'

Code:

public static void main(String[] args) {
    String temp="TTT..####";
    System.out.println(temp.length() - temp.replaceAll("T", "").length());
    System.out.println(temp.length() - temp.replaceAll(".", "").length());
    System.out.println(temp.length() - temp.replaceAll("#", "").length());
}

OutPut:

run:
3
9
4
BUILD SUCCESSFUL (total time: 1 second)

according to output '.' is 9 times in String.through loop it's gives the right answer. What is the problem??

Answer 1

Use replaceAll("\\\\.", "").length());

replaceAll() takes a REGEX as input. . has a special meaning in REGEX, it means any character . You need to escape it by using 2 \\\\ s

EDIT:

Use :

String temp = "TTT..####";
System.out.println(temp.split("\\.", -1).length - 1);

// using -1 as limit in split() divides the String based on the passed argument (and gives empty Strings as result, if needed). So, in the split array you will have n+1 elements, where n is the number of occurances of the particular argument. So, we are doing length-1 .

Answer 2

replaceAll() takes a regex and . mathces any char . This is the reason why your output is 9 in this case

Answer 3

You can find the number of times for every character in one iteration like this

Map<Character, Integer> map = new HashMap<>();
for(int i = 0; i < temp.length(); i++)
{
    char c = temp.charAt(i);
    if(!map.containsKey(c))
    {
        map.put(c, 1);
    }
    else
    {
        map.put(c, map.get(c)+1);
    }
}
for(Character c : map.keySet())
{
    System.out.println("" + c + " : " + map.get(c));
}

People have already answered that you need to use \\\\. instead of . in the regex. I just figured this could be a slightly more general solution for this particular problem with some minor alteration.