Java: Comparing ints and Strings - Performance

Question

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, s

Answer 1

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).

Answer 2

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

Answer 3

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.