两种方法求丑数

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。

方法1 ：

暴力破解，逐个判断

代码：

<pre name="code" class="cpp">#include <iostream>#include <vector>using namespace std;//判断是否是丑数bool isUgly(int index){while(index % 2 == 0){index /= 2;}while(index % 3 == 0){index /= 3;}while(index % 5 ==0){index /=5;}if(index == 1)return true;elsereturn false;}int  print(int index){int count=1;int number = 0;int uglyFound =0;while(uglyFound < index){++number;if(isUgly(number)){++uglyFound;}}return number;}int main(){cout<<print(1500);    return 0;}

运行结果：

方法2 ： 采用空间换时间，只是判断丑数。一个丑数可以有另外一个丑数* 2 或者*3 或者*５　得到。

#include <iostream>#include <vector>using namespace std;int Min(int pm2,int pm3,int pm5){int min = pm2 > pm3 ? pm3 : pm2;return min > pm5 ? pm5 : min;}void print(unsigned int index){if(index == 0)return;int * pUglyNumber = new int[index];int pNextIndex=1;pUglyNumber[0] = 1;int *pM2 = pUglyNumber;int *pM3 = pUglyNumber;int *pM5 = pUglyNumber;while(pNextIndex < index){int min=Min(*pM2 * 2,*pM3 * 3, *pM5 * 5);pUglyNumber[pNextIndex] = min;while(*pM2 * 2 <=pUglyNumber[pNextIndex])++pM2;while(*pM3 * 3 <=pUglyNumber[pNextIndex])++pM3;while(*pM5 * 5 <= pUglyNumber[pNextIndex])++pM5;pNextIndex ++;}int ugly = pUglyNumber[pNextIndex - 1];delete [] pUglyNumber;cout<< ugly;}int main(){print(7);    return 0;}

运行结果：