Java: Comparing ints and Strings - Performance

有些话、适合烂在心里 提交于 2019-11-29 13:28:52

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.

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).

Alade Fehinti John

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
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