Super Ugly Number
描述
Write a function to find the n
-th super ugly number.
Super ugly numbers are positive numbers whose all prime factors are in the given prime list primes
of size k
. For example, [1, 2, 4, 7, 8, 13, 14, 16, 19, 26, 28, 32]
is the sequence of the first 12 super ugly numbers given primes = [2, 7, 13, 19]
of size 4.
Note:
- 1 is a super ugly number for any given
primes
. - The given numbers in
primes
are in ascending order. - 0 <
k
≤ 100, 0 <n
≤ 1000000, 0 <primes[i]
< 1000.
分析
这题是 Ugly Number II 的扩展。在"Ugly Number II"中,primes=[2,3,5]
,这题中primes
可以自由变化。
所以这题可以用"Ugly Number II"的思路解决。每次要从多个列表中选择最小的元素,我们可以维护一个大小为primes
长度的小根堆。
代码
// Super Ugly Number
// Time complexity: O(n), Space complexity: O(n)
public class Solution {
public int nthSuperUglyNumber(int n, int[] primes) {
final int[] nums = new int[n];
nums[0] = 1; // 1 is the first ugly number
final Queue<Node> q = new PriorityQueue<>();
for (int i = 0; i < primes.length; ++i) {
q.add(new Node(0, primes[i], primes[i]));
}
for (int i = 1; i < n; ++i) {
// get the min element and add to nums
Node node = q.peek();
nums[i] = node.val;
// update top elements
do {
node = q.poll();
node.val = nums[++node.index] * node.prime;
q.add(node); // push it back
// prevent duplicate
} while (!q.isEmpty() && q.peek().val == nums[i]);
}
return nums[n - 1];
}
static class Node implements Comparable<Node> {
private int index;
private int val;
private int prime;
public Node(int index, int val, int prime) {
this.index = index;
this.val = val;
this.prime = prime;
}
public int compareTo(Node other) {
return this.val - other.val;
}
}
}