CODEWITHSUNDEEP
javaperformancecomparisonparseint
Justin
Justin

Reputation: 25387

Java: Comparing ints and Strings - Performance

I have a String and an int, lets say: String str = "12345"; and int num = 12345;. What is the fastest way of seeing if they are the same, str.equals("" + num) or num == Integer.parseInt(str) (Or is there a faster way?)?

This is the source code for Integer.parseInt and String.equals

Upvotes: 20

Views: 103763

Answers (3)

Alade Fehinti John
Alade Fehinti John

Reputation: 17

Guess you might also be able to use this to compare........

int p = 1234;
String Int = "1234";
String string = String.valueOf(p);
System.out.println(string + Int);
System.out.println(string.equals(Int));
code here

Upvotes: 0

dhruv chopra
dhruv chopra

Reputation: 498

num == Integer.parseInt(str) is going to faster than str.equals("" + num)

str.equals("" + num) will first convert num to string which is O(n) where n being the number of digits in the number. Then it will do a string concatenation again O(n) and then finally do the string comparison. String comparison in this case will be another O(n) - n being the number of digits in the number. So in all ~3*O(n)

num == Integer.parseInt(str) will convert the string to integer which is O(n) again where n being the number of digits in the number. And then integer comparison is O(1). So just ~1*O(n)

To summarize both are O(n) - but str.equals("" + num) has a higher constant and so is slower.

Upvotes: 23

Ankur Shanbhag
Ankur Shanbhag

Reputation: 7804

I think num == Integer.parseInt(str) is a better way of doing comparison. Because str.equals("" + num) this is not the ideal way how you should compare integer values and also it will create unnecessary String constant objects in the String pool (which hampers performance).

Upvotes: 5

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