返回字符串“hello/HELLO/hello...”出现在给定字符串中任意位置的次数

Question

Return the number of times the string "hello/Hello/...etc" appears anywhere in the given string.

The different in the problem is that

The string hello can be in any case ie either upper case or lower case.

Sample Input #1
count("abc hello def")

Sample Output #1
1

Sample Input #2
count("Hi. Hello. Hello. Ok")

Sample Output #2
2

Sample Input #3
count("hi")

Sample Output #3
0

MyApproach

public int count(String str)
    {
      String str1="Hello";

        int l=str.length();
        int l1=str1.length();
        if(l<l1)
        {
         return 0;
        }
        else
        {
           int count=0;
           int p=0;
           int j=0;
           while(j<l)
           {
            char c=str.charAt(j);
            char c1=str1.charAt(p);
            if(c==c1)
             {
               p++;

               if(p==l1)
               {
                count++;
                p=0;
               }  

             }
             else
             {
                 p=0;
             }
             j++;

           } 

     return count;
    }

}

Output       TestcaseParameters Testcase Actual Answer  Expected

No output    'HELLO how are you'               0             1

I am getting the following output.

Can anyone tell me what I am doing wrong?

Answer 1

I'm wondering how your code compiles. c1 is not even declared. I would probably use such logic:

str1.toUpperCase().split(str2.toUpperCase).length()-1

The code above does not return correct answer when str1 beginsWith or endsWith str2 . Here is a better version:

public static int count(String str1, String str2) {
    int count = 0;
    int len1 = str1.length();
    int len2 = str2.length();
    for (int i=0;i<=len1-len2;++i){
        if ((str1.substring(i,i+len2)).equalsIgnoreCase(str2)) {
            ++count;
        }
    }
    return count;
}

The code moves str2 rightwards once a step, and compares if str2 is the same as the substring of str1 . Please note in the situation str1 = "AAA", and str2 = "AA", the result is 2, since both the first and the second occurrences of AA will have a match.
Step 1:

str1: *****
str2: **

Step 2:

str1: *****
str2:  **

Step 3:

str1: *****
str2:   **

Step 4:

str1: *****
str2:    **

If you wish in the above situation, AA only counts once, here is the code for you:

public static int count(String str1, String str2) {
    int count = 0;
    int len1 = str1.length();
    int len2 = str2.length();
    for (int i=0;i<=len1-len2;){
        if ((str1.substring(i,i+len2)).equalsIgnoreCase(str2)) {
            ++count;
            i+=len2;
        }else{
            ++i;
        }

    }
    return count;
}

This version above works this way:

Step 1:

str1: *****
str2: **

Step 2:

str1: *****
str2:   **

It will not compare any character already compared in the previous steps.

I tested the code against your sample input data, and they all return the expected results.

Answer 2

If this is a school assignment then pls say so. Otherwise You will need to ask a question on google like: 'count occurrences in string java' or something. This will give you - tada - stackoverflow:

Occurrences of substring in a string

Hope to be of help.

Answer 3

Here is a pretty intuitive solution :

public static void main(String[] args) {
    System.out.println(count("hellohellohellohihihihihihihellohihhihello")); // Prints : 5
}

public static int count(String str){
    String str1 = "hello";
    String str2 = str.toLowerCase();
    int index = 0;
    int count = 0;

    while (true){
        if (str2.contains(str1)){
            count++;
            str2 = str2.substring(str2.indexOf(str1) + str1.length(), str2.length());
        } else {
            return count;
        }
    }
}

Answer 4

//in this we have changed every upper case character in lowercase for comparing it with s2 public int count(String str){ String s1="";String s2="hello"; int l2=s2.length(); int count=0;

    int l1=str.length();
    for(int i=0;i<=l1-l2;i++){
        for(int j=i;j<l2+i;j++){
            char ch=str.charAt(j);
            if(ch>=65&&ch<=90){
                ch=(char)(ch+32);
                s1=s1+ch;
            }
            else
            s1=s1+ch;
        }
        if(s1.equals(s2)){
            count++;
        }
//in that line i have removed the data of string s1 to form new string again
           s1="";
    }return count;

}