寻找丑数

题目：我们把只包含因子2、3和5的数称作丑数(Ugly Number)。例如6、8都是丑数，但14不是，因为它包含因子7。习惯上我们把1当做是第一个丑数。求按从小到大的顺序的第1500个丑数。

答：

#include "stdafx.h"
#include <iostream>

using namespace std;

//1、从1开始穷举，直到第1500个丑数，效率太低
//判断number是否为丑数
bool IsUgly(int number)
{
    while (number % 2 == 0)
    {
        number = number / 2;
    }
    while (number % 3 == 0)
    {
        number = number / 3;
    }
    while (number % 5 == 0)
    {
        number = number / 5;
    }
    return number == 1;
}

//寻找第index个丑数
int FindUglyOne(int index)
{
    if (index <= 0)
    {
        return -1;
    }
    int count = 0;
    int number = 0;
    while (count < index)
    {
        number++;
        if (IsUgly(number))
        {
            count++;
        }
    }
    return number;
}

int MinNumber(int a, int b, int c)
{
    a = a < b ? a : b;
    a = a < c ? a : c;
    return a;
}

//2、从丑数递推生成其他丑数
long FindUglyTwo(int index)
{
    if (index <= 0)
    {
        return -1;
    }
    int *arr = new int[index];
    arr[0] = 1;
    int index2 = 0;
    int index3 = 0;
    int index5 = 0;
    int count = 1;
    int ugly;
    while (count < index)
    {
        ugly = MinNumber(arr[index2] * 2, arr[index3] * 3, arr[index5] * 5);
        arr[count] = ugly;
        while(arr[index2] * 2 <= ugly)
        {
            index2++;
        }
        while(arr[index3] * 3 <= ugly)
        {
            index3++;
        }
        while(arr[index5] * 5 <= ugly)
        {
            index5++;
        }
        count++;
    }
    ugly = arr[count - 1];
    delete [] arr;
    return ugly;
}

int _tmain(int argc, _TCHAR* argv[])
{
    cout<<FindUglyOne(1500)<<endl;
    cout<<FindUglyTwo(1500)<<endl;
    return 0;
}

运行界面如下：