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. (4) The nth super ugly number is guaranteed to fit in a 32-bit signed integer.
就是找第n个超丑数
可以用动态规划
public int nthSuperUglyNumber(int n, int[] primes) { // write your code here if (n == 1) { return 1; } int[] num = new int[n]; int[] size = new int[primes.length]; num[0] = 1; for (int i = 1; i < n; i++) { int min = Integer.MAX_VALUE; for (int j = 0; j < size.length; j++) { if (min > num[size[j]] * primes[j]) { min = num[size[j]] * primes[j]; } } for (int j = 0; j < size.length; j++) { if (min == num[size[j]] * primes[j]) { size[j]++; } } num[i] = min; } return num[n - 1]; }可以用最小堆处理
public int nthSuperUglyNumber(int n, int[] primes) { // write your code here if (n == 1) { return 1; } Queue<Long> q = new PriorityQueue<Long>(); q.offer(1L); while (true) { long min = q.poll(); if (n == 1) { return (int)min; } for (int i = 0; i < primes.length; i++) { long cur = min * primes[i]; if (!q.contains(cur)) { q.offer(cur); } } n--; }}