// 面試題49：醜數
// 題目：我們把只包含因子2、3和5的數稱作醜數（Ugly Number）。求按從小到
// 大的順序的第1500個醜數。例如6、8都是醜數，但14不是，因為它包含因子7。
// 習慣上我們把1當做第一個醜數。

#include <iostream>

// ====================算法1的代碼====================
//不用額外的內存，直接計算
bool IsUgly(int number)//判斷是不是醜數
{
    while (number % 2 == 0)
        number /= 2;
    while
 
 (number % 3 == 0)
        number /= 3;
    while (number % 5 == 0)
        number /= 5;

    return (number == 1) ? true : false;
}

int GetUglyNumber_Solution1(int index)
{
    if (index <= 0)
        return 0;

    int number = 0;
    int uglyFound = 0;
    while (uglyFound < index)//從頭到尾開始計算
 

    {
        ++number;

        if (IsUgly(number))
            ++uglyFound;
    }

    return number;
}

// ====================算法2的代碼====================
//使用內存，只計算醜數，節省時間
int Min(int number1, int number2, int number3);

int GetUglyNumber_Solution2(int index)
{
    if (index <= 0)
        
 
return 0;

    int *pUglyNumbers = new int[index];
    pUglyNumbers[0] = 1;
    int nextUglyIndex = 1;

    int *pMultiply2 = pUglyNumbers;//用來存儲前面醜數的二倍值中，剛好大於當前最大醜數的數值
    int *pMultiply3 = pUglyNumbers;//三倍
    int *pMultiply5 = pUglyNumbers;//五倍

    while (nextUglyIndex < index)
    {
        int min = Min(*pMultiply2 * 2, *pMultiply3 * 3, *pMultiply5 * 5);//將這三個值中最小值作為下一個醜數存儲
        pUglyNumbers[nextUglyIndex] = min;

        while (*pMultiply2 * 2 <= pUglyNumbers[nextUglyIndex])//更新這三個值
            ++pMultiply2;
        while (*pMultiply3 * 3 <= pUglyNumbers[nextUglyIndex])
            ++pMultiply3;
        while (*pMultiply5 * 5 <= pUglyNumbers[nextUglyIndex])
            ++pMultiply5;

        ++nextUglyIndex;
    }

    int ugly = pUglyNumbers[nextUglyIndex - 1];
    delete[] pUglyNumbers;
    return ugly;
}

int Min(int number1, int number2, int number3)
{
    int min = (number1 < number2) ? number1 : number2;
    min = (min < number3) ? min : number3;

    return min;
}

// ====================測試代碼====================
void Test(int index, int expected)
{
    if (GetUglyNumber_Solution1(index) == expected)
        printf("solution1 passed\n");
    else
        printf("solution1 failed\n");

    if (GetUglyNumber_Solution2(index) == expected)
        printf("solution2 passed\n");
    else
        printf("solution2 failed\n");
}

int main(int argc, char* argv[])
{
    Test(1, 1);

    Test(2, 2);
    Test(3, 3);
    Test(4, 4);
    Test(5, 5);
    Test(6, 6);
    Test(7, 8);
    Test(8, 9);
    Test(9, 10);
    Test(10, 12);
    Test(11, 15);

    Test(1500, 859963392);

    Test(0, 0);
    system("pause");
    return 0;
}

《劍指offer》第四十九題（醜數）