Find a specific char in an array of chars

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 charac开发者_运维问答ter 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?

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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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You can use hash/map to check existance of the characer. This approach has better time of O(log n) or O(1) depending on hash/map inner structure.

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If you don't have access to Arrays, since you are working in JavaME, then you should try to:

• Or implement a sorted array and the binary search youself
• Or just use a O(n) solution, wich is a good solution anyways.

Your solution is O(n), along with the one stated by aix.

You could try to use a Map, but it would depends on how much elements you are in your array. If you think there won't be more than 1000 elements in the array, just use a O(n) solution. But if you think you could have an unkown number of elements, a Map would be a reasoable choice, providing a better solution.

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You can use net.rim.device.api.util.Arrays.getIndex(char[] array, char element)

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It depends on what Java ME configuration / profile you are using. If you're on CDC then check for what parts of Java SE 1.3 Collections framework are supported (just find javadocs targeted for your device and look in java.util package). Another thing worth checking is whether your device has some blackberry-specific API extensions to work with collections.

If you are limited to bare minimum of CLDC/MIDP then other solution than you mentioned would be to add chars to Vector and use Vector.contains(Object)

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If you don't want to implement it yourself you could use ArrayUtils in apache commons project:

ArrayUtils apache-commons