leetcode 313. Super Ugly Number 超级丑数

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 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.

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

这道题不太懂，还需要进一步思考。

建议和上一道题leetcode 263. Ugly Number 、 leetcode 264. Ugly Number II一起学习，

代码如下：

import java.util.Arrays;

/*
* http://www.cnblogs.com/grandyang/p/5144918.html *
* 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.
* */
public class Solution
{
public int nthSuperUglyNumber(int n, int[] primes)
{
int[] dp = new int
;
dp[0] = 1;
int[] idxPrimes = new int[primes.length];
int counter = 1;
while (counter < n)
{
int min = Integer.MAX_VALUE;
for (int i = 0; i < primes.length; i++)
{
// idxPrimes[i]代表每个丑数的个数，
// 比如丑数2题目的2，3，5，
// idxPrimes[0]代表2的下标
// idxPrimes[1]代表3的下标
// idxPrimes[2]代表5的下标
int temp = dp[idxPrimes[i]] * primes[i];
min = Math.min(min, temp);
}

// 如果min和 dp[idxPrimes[i]] * primes[i]相等，
// 则其对应的下标++
for (int i = 0; i < primes.length; i++)
{
if (min == dp[idxPrimes[i]] * primes[i])
idxPrimes[i]++;
}
dp[counter++] = min;
}
return dp[n - 1];
}
}

下面是C++的做法，就是一个简单的DP，我个人是不出来这种做法的，但是这个方法真的很棒

代码如下：

#include <iostream>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <string>
#include <climits>
#include <algorithm>
#include <sstream>

using namespace std;

class Solution
{
public:
int nthSuperUglyNumber(int n, vector<int>& primes)
{
vector<int> dp(n, 0);
dp[0] = 1;
vector<int> index(primes.size(),0);
int count = 1;
while (count < n)
{
int minNum = numeric_limits<int>::max();
for (int i = 0; i < primes.size(); i++)
{
int tmp = dp[index[i]] * primes[i];
minNum = min(minNum, tmp);
}

for (int i = 0; i < primes.size(); i++)
{
if (dp[index[i]] * primes[i] == minNum)
index[i]++;
}
dp[count++] = minNum;
}
return dp[n - 1];
}

int nthSuperUglyNumberBySet(int n, vector<int>& primes)
{
vector<int> res;
set<int> one{ 1 };
while (res.size() < n)
{
int minNum = *(one.begin());
one.erase(one.begin());
for (int p : primes)
one.insert(minNum*p);
res.push_back(minNum);
}
return res[n - 1];
}
};