Friday, January 8, 2016

LeetCode 326 - Power of Three


http://www.hrwhisper.me/leetcode-power-of-three/
Given an integer, write a function to determine if it is a power of three.
Follow up:
Could you do it without using any loop / recursion?
    bool isPowerOfThree(int n) {
        if(n <= 0) return false;
        while(n > 1){
            if(n %3 != 0) return false;
             n/=3;
        }
        return true;
    }
bool isPowerOfThree(int n) {
if (n <= 0) return false;
if (n == 1) return true;
return n % 3 == 0 && isPowerOfThree(n / 3);
}

bool isPowerOfThree(int n) {
return n <= 0 ? false : n == pow(3, round(log(n) / log(3)));
}
http://algobox.org/power-of-three/
A trivial loop / recursion solution is to divide the number by 3 and look at the remainder again and again.
Another idea is based on logarithm rules and float numbers which i don’t like very much. After all the log function it self may actually implemented using loop / recursion.
Since the 32 bit signed integer has at most 20 numbers, here is a list.
If we put them in a hash set beforehand, then we can query is power of three in O(1) time. I have seen people write in python like this
Python
Although it seems not using any loop, the in operation for list in python takes linear time, which indicating a loop underneath. We can use a set instead and put it into a class variable so we can share between calls.
Python
However, the fastest solution is this
Java


This works because 3 is a prime number. A 32 bit positive integer is a power of 3 is equivalent to it is a factor of 3^19 = 1162261467.
This doesn’t work for composite number like 4 or 6.
http://www.cnblogs.com/grandyang/p/5138212.html
题目中的Follow up让我们不用循环,那么有一个投机取巧的方法,由于输入是int,正数范围是0-231,在此范围中允许的最大的3的次方数为319=1162261467,那么我们只要看这个数能否被n整除即可,参见代码如下:
解法二:
class Solution {
public:
    bool isPowerOfThree(int n) {
        return (n > 0 && 1162261467 % n == 0);
    }
};

最后还有一种巧妙的方法,利用对数的换底公式来做,高中学过的换地公式为logab = logcb / logca,那么如果n是3的倍数,则log3n一定是整数,我们利用换底公式可以写为log3n = log10n / log103,注意这里一定要用10为底数,不能用自然数或者2为底数,否则当n=243时会出错,原因请看这个帖子。现在问题就变成了判断log10n / log103是否为整数,在c++中判断数字a是否为整数,我们可以用 a - int(a) == 0 来判断,参见代码如下:
解法三:
class Solution {
public:
    bool isPowerOfThree(int n) {
        return (n > 0 && int(log10(n) / log10(3)) - log10(n) / log10(3) == 0);
    }
};
http://fujiaozhu.me/?p=799
    bool isPowerOfThree(int n) {
        if(n == 0) return false;
        else return log10(n)/log10(3) - (int)(log10(n)/log10(3)) < 1e-15;
    }


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