If you don't mind creating an extra string:
String zeros = new String(new char[n]).replace((char) 0, '0');
Or more explicit (and probably more efficient):
char[] c = new char[n];
Arrays.fill(c, '0');
String zeros = new String(c);
Performance wise, the Arrays.fill
option seems to perform better in most situations, but especially for large strings. Using a StringBuilder
is quite slow for large strings but efficient for small ones. Using replace
is a nice one liner and performs ok for larger strings, but not as well as filll
.
Micro benchmark for different values of n:
Benchmark (n) Mode Samples Score Error Units
c.a.p.SO26504151.builder 1 avgt 3 29.452 ± 1.849 ns/op
c.a.p.SO26504151.builder 10 avgt 3 51.641 ± 12.426 ns/op
c.a.p.SO26504151.builder 1000 avgt 3 2681.956 ± 336.353 ns/op
c.a.p.SO26504151.builder 1000000 avgt 3 3522995.218 ± 422579.979 ns/op
c.a.p.SO26504151.fill 1 avgt 3 30.255 ± 0.297 ns/op
c.a.p.SO26504151.fill 10 avgt 3 32.638 ± 7.553 ns/op
c.a.p.SO26504151.fill 1000 avgt 3 592.459 ± 91.413 ns/op
c.a.p.SO26504151.fill 1000000 avgt 3 706187.003 ± 152774.601 ns/op
c.a.p.SO26504151.replace 1 avgt 3 44.366 ± 5.153 ns/op
c.a.p.SO26504151.replace 10 avgt 3 51.778 ± 2.959 ns/op
c.a.p.SO26504151.replace 1000 avgt 3 1385.383 ± 289.319 ns/op
c.a.p.SO26504151.replace 1000000 avgt 3 1486335.886 ± 1807239.775 ns/op