Returns the nth number of ugly

The numbers that contain only 2, 3, and 5 are called ugly Numbers (Ugly number), for example: 2,3,4,5,6,8,9,10,12,15, and so on, we used to think of 1 as the first ugly number.

Write an efficient algorithm to return the nth ugly number.

Solution 1: To determine whether a number is an ugly number, consecutive find the nth number of ugly

#include <iostream>
using namespace std;

BOOL isugly (int num) {
	if (num < 1) return
		false;
	if (num = 1) return
		true;
	while (num% 2 = 0)
		num/= 2;
	while (num% 3 = 0)
		num/= 3;
	while (num% 5 = 0)
		num/= 5;
	return num = = 1;
}

int nthugly (int n) {
	int index = 1, x = 2;
	while (Index < n) {
		if (isugly (x)) {
			index++;
			++x;
		}
		else
			++x;
	}
	return x-1;
}

int main () {
	int n;
	cout << "Input the number n:\n";
	CIN >> N;
	cout << nthugly (n);
	return 0;
}

Solution 2: Multiply 2,3,5 in the current ugly number to find the smallest subsequent ugly number added to the sequence

#include <iostream>
using namespace std;

int minOf3 (int a, int b, int c) {
	int temp = (A < b a:b);
	Return (Temp < C. temp:c);
}

Long findugly (int n) {
	int *ugly = new Int[n];
	Ugly[0] = 1;
	int index2 = 0, index3 = 0, index5 = 0;
	int index = 1;
	while (Index < n) {
		int val = minOf3 (Ugly[index2] * 2, UGLY[INDEX3] * 3, UGLY[INDEX5] * 5);
		Ugly[index] = val;
		while (Val >= ugly[index2] * 2)
			++index2;
		while (Val >= ugly[index3] * 3)
			++index3;
		while (Val >= ugly[index5] * 5)
			++index5;
		index++;
	}
	cout <<	ugly[n-1];
	delete [] ugly;
	return 0;
}

int main () {
	int n;
	cout << "Input the number n:\n";
	CIN >> N;
	findugly (n);
	return 0;
}