面试题43：丑数

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方法一：逐个数进行判断是否为丑数，效率低下

代码如下：

#include "stdafx.h"#include <iostream>using namespace std;//判断一个数是否为丑数  1被认为是第一个丑数bool IsUglyNum(int nNum){if (nNum <= 0){return false;}while (nNum % 2 == 0){nNum /= 2;}while (nNum % 3 == 0){nNum /= 3;}while (nNum % 5 == 0){nNum /= 5;}return (nNum == 1) ? true : false;}//返回第nIndex个丑数int GetUglyNum(int nIndex){if (nIndex <= 0){return 0;}int nNum = 1;int nUglyNum = 0;//记录第nIndex个丑数int nCount = 0;//计数while (nCount < nIndex){if (IsUglyNum(nNum)){nCount++;}nNum++;}nUglyNum = nNum - 1;return nUglyNum;}int _tmain(int argc, _TCHAR* argv[]){cout << GetUglyNum(7) << endl;system("pause");return 0;}

方法二：空间换时间，用一个数组保存排好序的丑数

代码如下：

#include "stdafx.h"#include <iostream>using namespace std;int Min(int a, int b, int c){int nMin = (a > b) ? b : a;return (nMin > c) ? c : nMin;}//返回第nIndex个丑数int GetUglyNum(int nIndex){    if (nIndex <= 0)    {return 0;    }//申请一个数组空间存放丑数int *pUglyNum = new int[nIndex];pUglyNum[0] = 1;int *pUglyNum2 = pUglyNum;int *pUglyNum3 = pUglyNum;int *pUglyNum5 = pUglyNum;int nNextIndex = 1;int nMin = 0;while (nNextIndex < nIndex){        nMin = Min((*pUglyNum2) * 2, (*pUglyNum3) * 3, (*pUglyNum5) * 5);pUglyNum[nNextIndex] = nMin;while ((*pUglyNum2) * 2 <= pUglyNum[nNextIndex]){++pUglyNum2;}while ((*pUglyNum3) * 3 <= pUglyNum[nNextIndex]){++pUglyNum3;}while ((*pUglyNum5) * 5 <= pUglyNum[nNextIndex]){++pUglyNum5;}nNextIndex++;}int nUglyNum = pUglyNum[nIndex - 1];delete [] pUglyNum;return nUglyNum;}int _tmain(int argc, _TCHAR* argv[]){cout << GetUglyNum(1500) << endl;system("pause");return 0;}