Java: Comparing ints and Strings - Performance

Justin picture Justin · Apr 13, 2013 · Viewed 70.1k times · Source

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

Answer

dhruv chopra picture dhruv chopra · Apr 13, 2013

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.