Find a specific char in an array of characters
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Find a specific char in an array of characters
I am doing an application where I want to find a specific char in an array of chars. In other words, I have the following char array:
char[] charArray= new char[] {'\uE001', '\uE002', '\uE003', '\uE004', '\uE005', '\uE006', '\uE007', '\uE008', '\uE009'};
At some point, I want to check if the character '\uE002'
exists in the charArray
. My method was to make a loop on every character in the charArray
and find if it exists.
for (int z = 0 ; z < charArray; z ++) {
if (charArray[z] == myChar) {
//Do the work
}
}
Is there any other solution than making a char array and finding the character by looping every single char?
One option is to pre-sort charArray
and use Arrays.binarySearch(charArray, myChar)
. A non-negative return value will indicate that myChar
is present in charArray
.
char[] charArray = new char[] {'\uE001', '\uE002', '\uE003', '\uE004', '\uE005', '\uE006', '\uE007', '\uE008', '\uE009'};
Arrays.sort(charArray); // can be omitted if you know that the values are already sorted
...
if (Arrays.binarySearch(charArray, myChar) >= 0) {
// Do the work
}
edit An alternative that avoids using the Arrays
module is to put the characters into a string (at initialization time) and then use String.indexOf():
String chars = "\uE001...";
...
if (chars.indexOf(myChar) >= 0) {
// Do the work
}
This is not hugely different to what you're doing already, except that it requires less code.
If n
is the size of charArray
, the first solution is O(log n)
whereas the second one is O(n)
.
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