Integer.parseInt() 和 Integer.toString() 运行时

Question

Would the runtime of Integer.parseInt(String i) and Integer.toString(int i) both be O(n)?

Answer 1

Yes both of them Integer.parseInt("1000") and Integer.toString(1000) have time complexity O(N)

• The internal code of Integer.parseInt("1000") reads the the strings char by char and covert to decimal in while loop

• The internal code of Integer.toString(1000) reads the integers and convert every digit to char and stores in byte[] buf then creates new string from the byte array

Here is the code of Integer.parseInt() :

 int i = 0, len = s.length(); int limit = -Integer.MAX_VALUE; // some checks int multmin = limit / radix; int result = 0; while (i < len) { // Accumulating negatively avoids surprises near MAX_VALUE int digit = Character.digit(s.charAt(i++), radix); if (digit < 0 || result < multmin) { throw NumberFormatException.forInputString(s, radix); } result *= radix; if (result < limit + digit) { throw NumberFormatException.forInputString(s, radix); } result -= digit; } return negative? result: -result;

Answer 2

Well, Thinking about it you can bypass the O(n) for Integer.toString(int i) by simply adding + ""

eg

String x = 555 + "";