313. Super Ugly Number

 

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 sizek. 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.

 

这道题让我们求超级丑陋数,是之前那两道Ugly Number 丑陋数Ugly Number II 丑陋数之二的延伸,质数集合可以任意给定,这就增加了难度。但是本质上和Ugly Number II 丑陋数之二没有什么区别,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,参见代码如下:

解法一:

class Solution {
public:
    int nthSuperUglyNumber(int n, vector<int>& primes) {
        vector<int> res(1, 1), idx(primes.size(), 0);
        while (res.size() < n) {
            vector<int> tmp;
            int mn = INT_MAX;
            for (int i = 0; i < primes.size(); ++i) {
                tmp.push_back(res[idx[i]] * primes[i]);
            }
            for (int i = 0; i < primes.size(); ++i) {
                mn = min(mn, tmp[i]);
            }
            for (int i = 0; i < primes.size(); ++i) {
                if (mn == tmp[i]) ++idx[i];
            }
            res.push_back(mn);
        }
        return res.back();
    }
};

 

上述代码可以稍稍改写一下,变得更简洁一些,原理完全相同,参见代码如下:

 

解法二:

class Solution {
public:
    int nthSuperUglyNumber(int n, vector<int>& primes) {
        vector<int> dp(n, 1), idx(primes.size(), 0);
        for (int i = 1; i < n; ++i) {
            dp[i] = INT_MAX;
            for (int j = 0; j < primes.size(); ++j) {
                dp[i] = min(dp[i], dp[idx[j]] * primes[j]);
            }
            for (int j = 0; j < primes.size(); ++j) {
                if (dp[i] == dp[idx[j]] * primes[j]) {
                    ++idx[j];
                }
            }
        }
        return dp.back();
    }
};

 

类似题目:

Ugly Number II

Ugly Number

 

参考资料:

https://leetcode.com/discuss/72835/108ms-easy-to-understand-java-solution

 

LeetCode All in One 题目讲解汇总(持续更新中…)


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