We can use the finditer() function inside the re module of Python for our specific problem. The finditer() function takes the pattern and the string as input parameters. It reads the string from left to right and returns all the indexes where the pattern occurs.
Use the string. count() Function to Find All Occurrences of a Substring in a String in Python. The string. count() is an in-built function in Python that returns the quantity or number of occurrences of a substring in a given particular string.
In order to find occurence of each character in a string we can use Map utility of Java.In Map a key could not be duplicate so make each character of string as key of Map and provide initial value corresponding to each key as 1 if this character does not inserted in map before.
A string is a collection of characters nested in double quotes. The indexOf method returns the index position of a specified character or substring in a string.
This should print the list of positions without the -1
at the end that Peter Lawrey's solution has had.
int index = word.indexOf(guess);
while (index >= 0) {
System.out.println(index);
index = word.indexOf(guess, index + 1);
}
It can also be done as a for
loop:
for (int index = word.indexOf(guess);
index >= 0;
index = word.indexOf(guess, index + 1))
{
System.out.println(index);
}
[Note: if guess
can be longer than a single character, then it is possible, by analyzing the guess
string, to loop through word
faster than the above loops do. The benchmark for such an approach is the Boyer-Moore algorithm. However, the conditions that would favor using such an approach do not seem to be present.]
Try the following (Which does not print -1 at the end now!)
int index = word.indexOf(guess);
while(index >= 0) {
System.out.println(index);
index = word.indexOf(guess, index+1);
}
String string = "bannanas";
ArrayList<Integer> list = new ArrayList<Integer>();
char character = 'n';
for(int i = 0; i < string.length(); i++){
if(string.charAt(i) == character){
list.add(i);
}
}
Result would be used like this :
for(Integer i : list){
System.out.println(i);
}
Or as a array :
list.toArray();
This can be done in a functional way with Java 9 using regular expression:
Pattern.compile(Pattern.quote(guess)) // sanitize input and create pattern
.matcher(word) // create matcher
.results() // get the MatchResults, Java 9 method
.map(MatchResult::start) // get the first index
.collect(Collectors.toList()) // collect found indices into a list
);
Here's the Kotlin Solution to add this logic as a new a new methods into CharSequence
API using extension method:
// Extension method
fun CharSequence.indicesOf(input: String): List<Int> =
Regex(Pattern.quote(input)) // build regex
.findAll(this) // get the matches
.map { it.range.first } // get the index
.toCollection(mutableListOf()) // collect the result as list
// call the methods as
"Banana".indicesOf("a") // [1, 3, 5]
With Java9, one can make use of the iterate(int seed, IntPredicate hasNext,IntUnaryOperator next)
as follows:-
List<Integer> indexes = IntStream
.iterate(word.indexOf(c), index -> index >= 0, index -> word.indexOf(c, index + 1))
.boxed()
.collect(Collectors.toList());
System.out.printlnt(indexes);
int index = -1;
while((index = text.indexOf("on", index + 1)) >= 0) {
LOG.d("index=" + index);
}
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