LeetCode 313. Super Ugly Number（超级丑的数字）

原题网址：https://leetcode.com/problems/super-ugly-number/

Write a program to find the nth 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) 1 is a super ugly number for any given primes.
(2) The given numbers in primes are in ascending order.
(3) 0 < k ≤ 100, 0 < n ≤ 106, 0 < primes[i] < 1000.

使用最小堆的最直接的思路，但无法通过时限要求：

public class Solution {    public int nthSuperUglyNumber(int n, int[] primes) {        PriorityQueue<Long> heap = new PriorityQueue<>();        for(int prime : primes) heap.offer((long)prime);        long ugly = 1;        for(int i = 1; i < n; i++) {            long prev = ugly;            ugly = heap.poll();            while (ugly == prev) ugly = heap.poll();            for(int prime : primes) heap.offer(ugly * prime);        }        return (int)ugly;    }}

方法：动态规划+最小堆。

public class Solution {    public int nthSuperUglyNumber(int n, int[] primes) {        if (n<=1) return 1;        int[] nums = new int[n];        nums[0] = 1;        int[] pos = new int[primes.length];        PriorityQueue<Integer> minHeap = new PriorityQueue<>(new Comparator<Integer>() {            @Override            public int compare(Integer i1, Integer i2) {               return Integer.compare(nums[pos[i1]] * primes[i1], nums[pos[i2]] * primes[i2]);            }        });        for(int i=0; i<primes.length; i++) minHeap.add(i);        for(int i=1; i<n; i++) {            do {                int min = minHeap.poll();                nums[i] = nums[pos[min]] * primes[min];                pos[min] ++;                minHeap.offer(min);            } while (nums[i]==nums[i-1]);        }        return nums[n-1];    }}