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.
【抛砖】
和Ugly Number II极为相似,只不过这里将3个数扩展为k个数,解释核心代码的含义:
minProduct[i] = superNumbers[indexs[i]] * primes[i];primes[i]为第i个素数,minProduct[i]存放着以i为索引的素数为因子的最小乘积,另一个因子存放在以index[i]为索引的superNumbers中:
public int nthSuperUglyNumber(int n, int[] primes) { int[] superNumbers = new int[n]; int[] indexs = new int[primes.length]; int[] minProduct = new int[primes.length]; superNumbers[0] = 1; for (int i = 0; i < primes.length; i++) minProduct[i] = superNumbers[indexs[i]] * primes[i]; int count = 1; while (count < n) { int minPrime = 0x7fffffff; for (int i = 0; i < primes.length; i++) minPrime = Math.min(minPrime, minProduct[i]); superNumbers[count++] = minPrime; for (int i = 0; i < primes.length; i++) if (minProduct[i] == minPrime) minProduct[i] = superNumbers[++indexs[i]] * primes[i]; } return superNumbers[n - 1]; }83 / 83 test cases passed. Runtime: 26 ms Your runtime beats 87.95% of javasubmissions.
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