《剑指offer》第四十九题(丑数)

时间:2023-03-09 06:55:05
《剑指offer》第四十九题(丑数)
// 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 % == )
number /= ;
while (number % == )
number /= ;
while (number % == )
number /= ; return (number == ) ? true : false;
} int GetUglyNumber_Solution1(int index)
{
if (index <= )
return ; int number = ;
int uglyFound = ;
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 <= )
return ; int *pUglyNumbers = new int[index];
pUglyNumbers[] = ;
int nextUglyIndex = ; 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 * , *pMultiply3 * , *pMultiply5 * );//将这三个值中最小值作为下一个丑数存储
pUglyNumbers[nextUglyIndex] = min; while (*pMultiply2 * <= pUglyNumbers[nextUglyIndex])//更新这三个值
++pMultiply2;
while (*pMultiply3 * <= pUglyNumbers[nextUglyIndex])
++pMultiply3;
while (*pMultiply5 * <= pUglyNumbers[nextUglyIndex])
++pMultiply5; ++nextUglyIndex;
} int ugly = pUglyNumbers[nextUglyIndex - ];
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\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[])
{
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system("pause");
return ;
}