A simple fast solution in Java. Uses approach described by Anon..
Here TreeSet
is just a container capable of returning smallest element in it. (No duplicates stored.)
int n = 20;
SortedSet<Long> next = new TreeSet<Long>();
next.add((long) 1);
long cur = 0;
for (int i = 0; i < n; ++i) {
cur = next.first();
System.out.println("number " + (i + 1) + ": " + cur);
next.add(cur * 2);
next.add(cur * 3);
next.add(cur * 5);
next.remove(cur);
}
Since 1000th ugly number is 51200000, storing them in bool[]
isn’t really an option.
edit
As a recreation from work (debugging stupid Hibernate), here’s completely linear solution. Thanks to marcog for idea!
int n = 1000;
int last2 = 0;
int last3 = 0;
int last5 = 0;
long[] result = new long[n];
result[0] = 1;
for (int i = 1; i < n; ++i) {
long prev = result[i - 1];
while (result[last2] * 2 <= prev) {
++last2;
}
while (result[last3] * 3 <= prev) {
++last3;
}
while (result[last5] * 5 <= prev) {
++last5;
}
long candidate1 = result[last2] * 2;
long candidate2 = result[last3] * 3;
long candidate3 = result[last5] * 5;
result[i] = Math.min(candidate1, Math.min(candidate2, candidate3));
}
System.out.println(result[n - 1]);
The idea is that to calculate a[i]
, we can use a[j]*2
for some j < i
. But we also need to make sure that 1) a[j]*2 > a[i - 1]
and 2) j
is smallest possible.
Then, a[i] = min(a[j]*2, a[k]*3, a[t]*5)
.