【leetcode】313. Super Ugly Number【M】【14】-优快云博客

本文链接：https://blog.youkuaiyun.com/sscssz/article/details/51837489

本文介绍了一种算法，用于求解给定一系列质数条件下第N个超丑数。超丑数是指所有质因数都在给定质数列表中的正整数。

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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.

Credits:
Special thanks to @dietpepsi for adding this problem and creating all test cases.

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申请一个长度为n的数组res，用于从小到大顺序存储n个丑数，数组中的首项为1，即第一个丑数为1

设置三个变量idx2、idx3、idx5存储下标，初始值都为0

找出数组res[idx2]*2、res[idx3]*3、res[idx5]*5的最小值，最小值即为下一个丑数，同时更新最小值对应的下标，如果多个数字同时为最小值，则它们的下标都要更新

找到第n个丑数时，循环结束

class Solution(object):
    def nthSuperUglyNumber(self, n, primes):
        res = [1]
        ind = [0] * len(primes)

        for i in xrange(n+1):
            t = []

            for j in xrange(len(primes)):
                t += res[ind[j]] * primes[j],
            minn = min(t)
            for j in xrange(len(primes)):
                if t[j] == minn:
                    ind[j] += 1

            res += minn,
            #print t,res

        return res[n-1]