string integer compare java

Santino answered on February 27, 2021 Popularity 10/10 Helpfulness 2/10

answer string integer compare java

string integer compare java

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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.

Popularity 10/10 Helpfulness 2/10 Language java

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Contributed on Feb 27 2021

Santino

0 Answers Avg Quality 2/10

string integer compare java

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