• 《剑指offer》丑数


    本题来自《剑指offer》 丑数

    题目:

       把只包含质因子2、3和5的数称作丑数(Ugly Number)。例如6、8都是丑数,但14不是,因为它包含质因子7。 习惯上我们把1当做是第一个丑数。求按从小到大的顺序的第N个丑数。

    思路:

       两种思路:

      第一种:丑数的定义是只包含质因子2,3,5.那么该数如果一直能被235整除直到为1,那么就是丑数,python代码实现。该代码直观明了,用时间换空间。由于用到了多重循环,而且每个数字都要被计算,时间花销较大。

      第二种:用空间换时间.丑数应该是另一个丑数乘以2,3,5的结果。可以定一个一块空间,存储着已经排好序的丑数,然后在乘以2,3,5.这种思路的前提是丑数应该是排好序的。C++代码实现。

    C++ Code:

    class Solution {
    public:
        int GetUglyNumber_Solution(int index) {
            if (index<=0){                                                       //boundary condition judgement
                return 0;
            }
            int *pUglyNumbers = new int[index];                                  //cache
            pUglyNumbers[0] = 1;                                                 //the first number is 1
            int nextUglyIndex = 1;                                               //the index start from 1
            int *pMultiply2 = pUglyNumbers;
            int *pMultiply3 = pUglyNumbers;
            int *pMultiply5 = pUglyNumbers;
            while (nextUglyIndex<index){ 
                int min = Min(*pMultiply2*2,*pMultiply3*3,*pMultiply5*5);        //get the smallest of the three numbers to cache it
                pUglyNumbers[nextUglyIndex] = min;
                while(*pMultiply2*2<=pUglyNumbers[nextUglyIndex]){               
                    pMultiply2++;
                }
                while(*pMultiply3*3<=pUglyNumbers[nextUglyIndex]){
                    pMultiply3++;
                }
                while(*pMultiply5*5<=pUglyNumbers[nextUglyIndex]){
                    pMultiply5++;
                }
                nextUglyIndex++;
            }
            int ugly = pUglyNumbers[nextUglyIndex-1];                          //the last one in the cache is the result value
            delete[] pUglyNumbers;                                             //delete the cache,relative to new
            return ugly;
        }
        int Min(int num1,int num2,int num3){                                   //the smallest of the three numbers
            int min = (num1<num2)?num1:num2;
            min = (min<num3)?min:num3;
            return min;
        }
    };

    Python Code:

    # -*- coding:utf-8 -*-
    class Solution:
        def GetUglyNumber_Solution(self, index):
            # write code here
            result = 0
            if index >= 1:
                result = 1                    #the theory is that the first ugly number is 1,and the count start from 1
            for num in range(2,index+1):      #loop from 2 to the last number
                if self.IsUgly(num):
                    result += 1               #counter accumulation statistics
            return result 
        def IsUgly(self,num):                 #this program is to detmine whether the single number is ugly
            while(num%2 == 0):                #detmine that the number can be divisible by 2
                num /= 2
            while(num%5 == 0):                #detmine that the number can be divisible by 5
                num /= 5
            while(num%3 == 0):                #detmine that the number can be divisible by 3
                num /= 3
            if num == 1:                      #only when the result is 1,this number is ugly
                return True
            else:
                return False

    总结:

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  • 原文地址:https://www.cnblogs.com/missidiot/p/10783671.html
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