How can I compare two strings in java and define which of them is smaller than the other alphabetically?

Question

I want to use the binary search algorithm to search the string which has been entered by the user in a very big sorted file. I can not compare the string which has been ente

Answer 1

Haven't you heard about the Comparable interface being implemented by String ? If no, try to use

"abcda".compareTo("abcza")

And it will output a good root for a solution to your problem.

Answer 2

If you would like to ignore case you could use the following:

String s = "yip";
String best = "yodel";
int compare = s.compareToIgnoreCase(best);
if(compare < 0){
    //-1, --> s is less than best. ( s comes alphabetically first)
}
else if(compare > 0 ){
// best comes alphabetically first.
}
else{
    // strings are equal.
}

Answer 3

You can use

str1.compareTo(str2);

If str1 is lexicographically less than str2, a negative number will be returned, 0 if equal or a positive number if str1 is greater.

E.g.,

"a".compareTo("b"); // returns a negative number, here -1
"a".compareTo("a"); // returns  0
"b".compareTo("a"); // returns a positive number, here 1
"b".compareTo(null); // throws java.lang.NullPointerException