LeetCode: Super Ugly Number

最新推荐文章于 2024-06-04 11:57:06 发布

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本文介绍了一种高效算法来寻找第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。通过记录每个值对应的质数位置并利用最小值策略，避免了暴力枚举的高时间复杂度。

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 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.

题目解析:这是Ugly Number II 的拓展,

这里,最简单的方法就是枚举,然后一一判断,但这样时间代价过高.

比较简单的方法是记录每个值走选到了第几个素数,假设只有一个素数,那么res[i] = res[i- 1] * primes[j], 那么下一个素数res[i + 1] = res[i] * primes[j + 1]

多个素数的时候就需要记录下一个选取的素数是哪个.

源代码:

class Solution {
public:
    int nthSuperUglyNumber(int n, vector<int>& primes) {
        // if (n == 1) return 1;
        
        int len = primes.size();
        vector<int> id(len, 0);
        vector<int> res(n, 0);
        res[0] = 1;
        
        for (int i = 1; i < n; ++i)
        {
            int m = INT_MAX;
            for (int j = 0; j < len; ++j) {
                m = min(m, res[id[j]] * primes[j]);
            }
            
            res[i] = m;
            for (int j = 0; j < len; ++j) {
                if (res[id[j]] * primes[j] == m) {
                    id[j]++;
                }
            }
        }
        
        return res[n - 1];
        
    }
};