Geeks 面试题： Ugly Numbers

Ugly Numbers

Ugly numbers are numbers whose only prime factors are 2, 3 or 5. The sequence
1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, …
shows the first 11 ugly numbers. By convention, 1 is included.
Write a program to find and print the 150′th ugly number.

METHOD 1 (Simple)
Thanks toNedylko Draganovfor suggesting this solution.

Algorithm:
Loop for all positive integers until ugly number count is smaller than n, if an integer is ugly than increment ugly number count.

To check if a number is ugly, divide the number by greatest divisible powers of 2, 3 and 5, if the number becomes 1 then it is an ugly number otherwise not.

For example, let us see how to check for 300 is ugly or not. Greatest divisible power of 2 is 4, after dividing 300 by 4 we get 75. Greatest divisible power of 3 is 3, after dividing 75 by 3 we get 25. Greatest divisible power of 5 is 25, after dividing 25 by 25 we get 1. Since we get 1 finally, 300 is ugly number.

下面是这个简单方法，这里带了可以打印所有小于n的ugly number的程序：

int maxDivide(int num, int div)
{
	while (num % div == 0)
	{
		num /= div;
	}
	return num;
}

bool isUgly(int num)
{
	num = maxDivide(num, 2);
	num = maxDivide(num, 3);
	num = maxDivide(num, 5);
	return num == 1? true:false;
}

int getNthUglyNo(int n)
{
	int c = 0;
	int i = 0;
	while (c < n)
	{
		if (isUgly(++i)) c++;
	}
	return i;
}
#include <vector>
using std::vector;
vector<int> getAllUglyNo(int n)
{
	vector<int> rs;
	for (int i = 1; i <= n; i++)
	{
		if (isUgly(i)) rs.push_back(i);
	}
	return rs;
}

动态规划法：

注意：可能会有重复的数字要跃过，程序注释出去了。否则答案错误！

细想一下，其实一句话概括就是：

由最小的Ugly number算起，然后每个都分别乘以2,3,5得到最终答案。

class UglyNumbers
{
public:
	int getNthUglyNo(int n, vector<int> &rs)
	{
		if (n < 1) return n; 
		int n2 = 2, n3 = 3, n5 = 5;
		int i2 = 0, i3 = 0, i5 = 0;
		rs.resize(n, 1);
		for (int i = 1; i < n; i++)
		{
			int t = min(n2, min(n3,n5));
			if (t == n2) 
			{
				rs[i] = n2;
				n2 = rs[++i2]*2;
			}
			if (t == n3) //注意：不能使用else，避免重复
			{
				rs[i] = n3;
				n3 = rs[++i3]*3;
			}
			if (t == n5) //注意：不能使用else
			{
				rs[i] = n5;
				n5 = rs[++i5]*5;
			}
		}
		return rs.back();
	}
};

测试：

int main()
	{
		unsigned no = getNthUglyNo(35);
		printf("ugly no. is %d \n",  no);
		vector<int> rs = getAllUglyNo(35);
		for (auto x:rs) cout<<x<<" ";
		cout<<endl;

		UglyNumbers un;
		printf("Ugly no. is %d \n", un.getNthUglyNo(35, rs));
		for (auto x:rs) cout<<x<<" ";
		cout<<endl;

		system("pause");	
		return 0;
	}