For example, what would be the most efficient way to get say 999 if given an n that equals 3 for instance.
This is what I have got right now but I was wondering if there was a more elegant way.
public static int largestPossibleNumber(int numDigits) {
return Integer.parseInt(new String(new char[numDigits]).replace("\0", "9"));
}
Example Usage:
for (int i = 1; i <= 5; i++) {
System.out.println(largestPossibleNumber(i));
}
Output:
9
99
999
9999
99999
You have just 8 valid answers, so you can hardcode them:
private static int[] s_Numbers = {
0, 9, 99, 999, 9999, 99999, 999999, 9999999, 99999999};
private static int largestPossibleNumber(int n) {
return s_Numbers[n];
}
You are asking for the most efficient way. It's rather hard to prove that some way is the most efficient - it would require implementing and benchmarking multiple methods at least.
But here's a pretty fast way to do this — just create a Map, or use a switch, see below.
This works because the size of an int is fixed. Note however, that this method won't extend to, say, BigIntegers.
public static int largestPossibleNumber(final int numDigits) {
switch (numDigits) {
case 1: return 9;
case 2: return 99;
case 3: return 999;
case 4: return 9999;
case 5: return 99999;
case 6: return 999999;
case 7: return 9999999;
case 8: return 99999999;
case 9: return 999999999;
case 10: return Integer.MAX_VALUE;
default: throw new IllegalArgumentException();
}
}
public static int largestPossibleNumber(int n) {
return (int) (Math.pow(10.0, n)) -1;
}
public static int largestPossibleNumber(int numDigits) {
return (int) (Math.pow(10, numDigits)) - 1;
}