超级丑数 Super Ugly Number

LeetCode传送门313. Super Ugly Number

题目

超级丑数 是一个正整数，并满足其所有质因数都出现在质数数组 primes 中。

给你一个整数 n 和一个整数数组 primes ，返回第 n 个 超级丑数 。

题目数据保证第 n 个 超级丑数 在 32-bit 带符号整数范围内。

A super ugly number is a positive integer whose prime factors are in the array primes.

Given an integer n and an array of integers primes, return the nth super ugly number.

The nth super ugly number is guaranteed to fit in a 32-bit signed integer.

Example:

Input: n = 12, primes = [2,7,13,19]
Output: 32
Explanation: [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].

Input: n = 1, primes = [2,3,5]
Output: 1
Explanation: 1 has no prime factors, therefore all of its prime factors are in the array primes = [2,3,5].

Constraints:

1 <= n <= 10^6 1 <= primes.length <= 100 2 <= primes[i] <= 1000 primes[i] is guaranteed to be a prime number. All the values of primes are unique and sorted in ascending order.

思考线

解题思路

本道题是上一篇文章的升级版，也是用动态规划去解决，只不过我们小心的去处理primes中对应pd中的下标。

如果学会了上一篇 丑数II，这个部分的代码就呼之欲出，直接上代码

function nthSuperUglyNumber(n: number, primes: number[]): number {
    const pivot:number[] = new Array(primes.length).fill(1);
    const pd = [0,1];
    console.log(pivot)
    for(let i = 2; i <=n;i++) {
        let min = Infinity;

        const res = pivot.map((item, index) => {
             const r = pd[item] * primes[index];
             if(r<=min) {
                 min = r;
             }
             return r;
        });
        pd[i] = min;
        res.forEach((item, index) => {
            if(item === min) {
                pivot[index]++;
            }
        })
    }
    return pd[n];
};

这就是我对本题的解法，如果有疑问或者更好的解答方式，欢迎留言互动。