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 丑陋数和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(, ), idx(primes.size(), );
while (res.size() < n) {
vector<int> tmp;
int mn = INT_MAX;
for (int i = ; i < primes.size(); ++i) {
tmp.push_back(res[idx[i]] * primes[i]);
}
for (int i = ; i < primes.size(); ++i) {
mn = min(mn, tmp[i]);
}
for (int i = ; 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, ), idx(primes.size(), );
for (int i = ; i < n; ++i) {
dp[i] = INT_MAX;
for (int j = ; j < primes.size(); ++j) {
dp[i] = min(dp[i], dp[idx[j]] * primes[j]);
}
for (int j = ; j < primes.size(); ++j) {
if (dp[i] == dp[idx[j]] * primes[j]) {
++idx[j];
}
}
}
return dp.back();
}
};
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参考资料：
https://leetcode.com/discuss/72835/108ms-easy-to-understand-java-solution