Leetcode Note: Super uugly Number

Leetcode Note: Super uugly Number

I. Description

Write a program to find the nth super uugly number.

Super uugly 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 uugly numbers given primes = [2, 7, 13, 19] of size 4.

Note:

1 is a super uugly number for any given primes. the given numbers in primes are in ascending order. 0 <k ≤ 100, 0 <n ≤ 106, 0 <primes [I] <1000.

Ii. Question Analysis

The general idea is to write a program to find the nth "super ugly Number". The definition of the super ugly number is as follows:

The super ugly number refers to the positive number that only contains the given k quality factors. For example, given the prime number sequence with a length of 4, primes = [2, 7, 13, 19], the first 12 super ugly number sequences are: [1, 2, 4, 7, 8, 13, 14, 16, 19, 26, 28, 32]

Note:

1 is considered as a super ugly number, regardless of the given prime number list. The given prime numbers are listed in ascending order. 0 <k ≤ 100, 0 <n ≤ 106, 0 <primes [I] <1000.

One way is to use an array to record eachprimesNumber of times of the product of the inner prime number, which is stored in another Array1TokUgly number.

Iii. Sample Code

class Solution {  public:      int nthSuperUglyNumber(int n, vector
  
   & primes) {          int len = primes.size();          vector
   
     index(len, 0);        vector
    
      uglyNum(n, INT_MAX);        vector
     
       temp(len);        uglyNum[0] = 1;        for (int i = 1; i < n; ++i)        {            int minj = -1;            int minNum = INT_MAX;            for (int j = 0; j < len; ++j)            {                temp[j] = primes[j] * uglyNum[index[j]];                if (temp[j] < uglyNum[i])                {                    minNum = temp[j];                    uglyNum[i] = temp[j];                    minj = j;                }            }            for (int j = minj; j < len; ++j)            {                if (minNum == temp[j])                ++index[j];            }        }        return uglyNum[n - 1];    }  };
     
    
   
  

Iv. Summary

In fact, this method requires at least 10 lines of code. In order to omit some operations, more auxiliary memory is used.