p78 3的幂 (leetcode 326)

一：解题思路

方法一：可以采用前面讲解的类似的2的幂来做这道题，Time:O(log_3(n)),Space:O(1)

方法二：整数最大值，y=2^31-1。3^a<=y,那么a<=log_ay ==> a<=ln(y)/ln(3)，计算出a为19.所以MAX_NUM=3^19。Time:O(1),Space:O(1)

二：完整代码示例 (C++版和Java版)

方法一C++：

class Solution {
public:
    bool isPowerOfThree(int n) 
    {
        if (n <= 0) return false;
        while (n % 3 == 0)
            n /= 3;
        return n == 1;
    }
};

方法一Java:

class Solution {
        public boolean isPowerOfThree(int n) 
        {
               if(n<=0) return false;
               while(n%3==0)
               {
                   n/=3;
               }
               
               return n==1;
        }
    }

方法二C++:

class Solution {
public:
    int MAX_POWER = (int)(log(2147483647) / log(3));
    int MAX_NUM = (int)pow(3, MAX_POWER);

    bool isPowerOfThree(int n) 
    {
        return n > 0 && (MAX_NUM%n==0);
    }
};

方法二Java:

class Solution {
        private final int MAX_POWER=(int)(Math.log(Integer.MAX_VALUE)/Math.log(3));
        private final int MAX_NUM=(int)(Math.pow(3,MAX_POWER));
        public boolean isPowerOfThree(int n)
        {
               return n>0 && (MAX_NUM%n)==0;
        }
    }

标签：log,POWER,int,MAX,isPowerOfThree,p78,public,326,leetcode
来源： https://www.cnblogs.com/repinkply/p/12628937.html