# Simple java solution with explanation

• ``````public int nthUglyNumber(int n) {
// 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20.... is the sequence of the ugly numbers
// 2, 3, 2*2, 5, 2*3, 3*3, 2*5, 3*5
// 2^a 3^b 5^c
PriorityQueue<Integer> pq = new PriorityQueue<Integer>(new Comparator<Integer>() {
public int compare(Integer a, Integer b) {
return a - b;
}
});
for (int i = -1; i * i <= n; i++) {
for (int j = -1; j * j <= n; j++) {
for (int k = -1; k * k <= n; k++) {
pq.add((int)(Math.pow(5, i + 1)*Math.pow(3, j + 1)*Math.pow(2, k + 1)));
}
}
}
int res = 1;
for (int i = 0; i < n; i++) {
res = pq.poll();
}
return res;
}
``````

idea: every ugly number is comprised of multiple 2, times multiple 3, and multiple 5. That is ugly number = Math.pow(2, x)*Math.pow(3, y)*Math.pow(5, z). Consider the method of counting the primes. I use sqrt() here to reduce the useless calculation. Meantime, I use the heap to get the nth element.

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