Domanda di colloquio di Microsoft
Nothing really difficult.
Risposta di colloquio
Anonimo
20 feb 2013
int counter = 0; for (int i = a.length-1; i >= 0; i--) { if (a[i] != 0){ a[a.length-1-counter] = a[i]; a[i] = 0; counter++; } } O(n). One pass.
Nothing really difficult.
Anonimo
int counter = 0; for (int i = a.length-1; i >= 0; i--) { if (a[i] != 0){ a[a.length-1-counter] = a[i]; a[i] = 0; counter++; } } O(n). One pass.