《剑指offer》第四十九题（丑数）

Posted 2021-02-13 cjt-blog

// 面试题49：丑数
// 题目：我们把只包含因子2、3和5的数称作丑数（Ugly Number）。求按从小到
// 大的顺序的第1500个丑数。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当做第一个丑数。

#include <iostream>

// ====================算法1的代码====================
//不用额外的内存，直接计算
bool IsUgly(int number)//判断是不是丑数
{
    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)//从头到尾开始计算
    {
        ++number;

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

    return number;
}

// ====================算法2的代码====================
//使用内存，只计算丑数，节省时间
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;//用来存储前面丑数的二倍值中，刚好大于当前最大丑数的数值
    int *pMultiply3 = pUglyNumbers;//三倍
    int *pMultiply5 = pUglyNumbers;//五倍

    while (nextUglyIndex < index)
    {
        int min = Min(*pMultiply2 * 2, *pMultiply3 * 3, *pMultiply5 * 5);//将这三个值中最小值作为下一个丑数存储
        pUglyNumbers[nextUglyIndex] = min;

        while (*pMultiply2 * 2 <= pUglyNumbers[nextUglyIndex])//更新这三个值
            ++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;
}

// ====================测试代码====================
void Test(int index, int expected)
{
    if (GetUglyNumber_Solution1(index) == expected)
        printf("solution1 passed
");
    else
        printf("solution1 failed
");

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

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;
}