Count the number of given strings in a string

1. Split method

1.1 Thought analysis

For example, the string is "1java2java3e" With java To split strings

The resulting string array String[] strs={“1”,“2”,“3e”}, The length of 3

And the number of strings that meet the conditions =2= String array length -1, Therefore, we have the following conclusions :

The length of the split string -1= Number of given characters

1.2 Sample source code

package Work;
public class Test05 {
    
    public static void main(String[] args) {
    
        System.out.println(countString("1java23javajav", "java"));
        System.out.println(countString("zdhwefyg", "java"));
    }
    public static int countString(String str,String regex){
    
        //str Is the original string ,regex For the given string 
        return str.split(regex).length-1;
    }

}

1.3 Screenshot of sample source code

2. Replace the statistical method

2.1 Thought analysis

Set the original string to str1

Replace the given string existing in the original string with ""( Empty string ), Get a new string str2;

The number of given strings =(str1-str2)/ The length of a given string

2.2 Sample source code

package Work;
public class Test05 {
    
    public static void main(String[] args) {
    
        System.out.println(countString("java1java23javajav", "java"));
        System.out.println(countString("zdhwefyg", "java"));
    }
    public static int countString(String str,String regex){
    
        //str Is the original string ,regex For the given string 
        return (str.length()-str.replace(regex,"").length())/regex.length();
    }

}

2.3 Screenshot of sample source code

3. The subscript method

3.1 Thought analysis

adopt indexOf Find the subscript where the given string first appears index,

if -1, The statistical result is 0

If not -1, Go into the cycle ( The loop termination condition is indexOf The result is zero -1). here count The quantity needs to be increased 1

Then from the last character in the found string (index+regex.length) Start looking for a given string , If you have any , Do not exit the loop ,count Keep adding 1, Then continue to look back ( Looking for the position is still the same rule as the front ), Until there is no given string , So end the cycle , The final return is count The number of

3.2 Sample source code

package Work;
public class Test05 {
    
    public static void main(String[] args) {
    
        System.out.println(countString("java1java23javajav", "java"));
        System.out.println(countString("zdhwefyg", "java"));
    }
    public static int countString(String str,String regex){
    
        //str Is the original string ,regex For the given string 
        int index=str.indexOf(regex);
        // Find whether the original string has a given string for the first time , If -1( It means there is no ), Then there is no need to look back , If there is, enter the cycle to find 
        int count=0;
        while (index!=-1){
    
            // Enter the cycle count+1
            count++;
            // Then it needs to be updated every time index Value , Because you need to start from the last position of the string 
            // There is no need to look from the past to the future 
            index=str.indexOf(regex,index+regex.length());
        }
        return  count;
    }

}

3.3 Screenshot of sample source code

4. Subscript + Intercepting substring method

4.1 analysis

adopt indexOf Find the subscript where the given string first appears index,

If -1, The statistical result is 0, There is no need to intercept later

If not for -1, Just go into the cycle to find ( The condition of circulation is not -1),

Every time the cycle is established count+1, Then after it is established, you need to intercept from the last character of the given string ( Because the previous comparison is over , There is no need to intercept )

After interception You need to get the subscript of the given string in the intercepted substring , If not -1 Just keep cycling , otherwise

Exit loop , Then intercept like this all the time, and then add the subscript to get it , Until you get the subscript -1, Just exit the loop

4.2 Sample source code

package Work;
public class Test05 {
    
    public static void main(String[] args) {
    
        System.out.println(countString("j2ava1java23javajav", "java"));
        System.out.println(countString("zdhwefyg", "java"));
    }
    public static int countString(String str,String regex){
    
        //str Is the original string ,regex For the given string 
        int index=str.indexOf(regex);
        // Find whether the original string has a given string for the first time , If -1( It means there is no ), Then there is no need to look back , If there is, enter the cycle to find 
        int count=0;
        while (index!=-1){
    
            // Enter the cycle count+1
            count++;
           // Every time you do it, you need to intercept + Get the position of the given string in the substring after interception 
            str=str.substring(index+regex.length());
            //indexOf The default is from 0 Start , It just starts from the first bit of the string 
            index=str.indexOf(regex);
        }
        return  count;
    }

}

4.3 Screenshot of sample source code