Related questions
Make a code using Recursion
The countSubstring function will take two strings as parameters and will return an integer that is the count of how many times the substring (the second parameter) appears in the first string without overlapping with itself. This method will be case insensitive.
For example:
countSubstring("catwoman loves cats", "cat") would return 2
countSubstring("aaa nice", "aa") would return 1 because "aa" only appears once without overlapping itself.
public static int countSubstring(String s, String x) {
if (s.length() == 0 || x.length() == 0)
return 1;
if (s.length() == 1 || x.length() == 1){
if (s.substring(0,1).equals(x.substring(0,1))){
s.replaceFirst((x), " ");
return 1 + countSubstring(s.substring(1), x);
}
else {
return 0 + countSubstring(s.substring(1), x);
}
}
return countSubstring(s.substring(0,1), x) + countSubstring(s.substring(1), x);
}
public class Main {
public static void main(String[] args) {
System.out.println(Recursion.countSubstring("catwoman loves cats","cat"));
System.out.println(Recursion.countSubstring("aaa nice","aa"));
}
}
The source code of the program
#Main.java
class Main
{
public static void main(String[] args)
{
String testString3 = "catwoman loves cats";
String testString2 = "aaa nice";
System.out.println(Recursion.countSubstring(testString3, "cat"));
System.out.println(Recursion.countSubstring(testString2, "aa"));
}
}
#Recursion.java
class Recursion
{
public static int countSubstring(String s, String sub)
{
int subLength = sub.length();
int sLength = s.length();
if (sLength < subLength)
{
return 0;
}
else
{
String sLower = s.toLowerCase();
String subLower = sub.toLowerCase();
if (sLower.substring(0, subLength).equals(subLower))
{
return 1 + countSubstring(s.substring(subLength), sub);
}
else
{
return countSubstring(s.substring(1), sub);
}
}
}
}
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