LeetCode-313. Super Ugly Number (JAVA)超级丑数

313. Super Ugly Number

Write a program to find the n^th super ugly number.

Super ugly numbers are positive numbers whose all prime factors are in the given prime listprimes 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 givenprimes = [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 n^th super ugly number is guaranteed to fit in a 32-bit signed integer.

与丑数不同的是，之前求丑数是从2，3，5中找，而本题是给定的质数，求出他们的倍数的超级丑数

质数集合可以任意给定，由于我们不知道质数的个数，我们可以用一个idx数组来保存当前的位置，然后我们从每个子链中取出一个数，找出其中最小值，然后更新idx数组对应位置，注意有可能最小值不止一个，要更新所有最小值的位置

参考：

 LeetCode-263.264. Ugly Number  (JAVA)丑数

public int nthSuperUglyNumber(int n, int[] primes) {int[] dp = new int[n];// 第一个超级丑数是1dp[0] = 1;int[] idxPrimes = new int[primes.length];int counter = 1;while (counter < n) {int min = Integer.MAX_VALUE;for (int i = 0; i < primes.length; i++) {// idxPrimes[i]代表每个丑数的个数，// 比如丑数2题目的2，3，5，// idxPrimes[0]代表2的下标// idxPrimes[1]代表3的下标// idxPrimes[2]代表5的下标int temp = dp[idxPrimes[i]] * primes[i];min = min < temp ? min : temp;}// 如果min和 dp[idxPrimes[i]] * primes[i]相等，// 则其对应的下标++for (int i = 0; i < primes.length; i++) {if (min == dp[idxPrimes[i]] * primes[i]) {idxPrimes[i]++;}}dp[counter] = min;counter++;}return dp[n - 1];}