[LeetCode]114. Power of Three 三的幂

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?

解法1：不断除去n的因子3，最后判断是否为1

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

解法2：将循环改为递归。

class Solution {
public:
    bool isPowerOfThree(int n) {
        if(n != 0 && n % 3 == 0) {
            n /= 3;
            return isPowerOfThree(n);
        }
        else if(n == 1) return true;
        else return false;
    }
};

解法3：不使用循环或递归。

class Solution {
public:
    bool isPowerOfThree(int n) {
        return fmod(log10(n) / log10(3), 1) == 0;
    }
};

因为n限制为int类型，最大为3^19=1162261467，所以只要判断在n不为0时是否能整除1162261467即可。

class Solution {
public:
    bool isPowerOfThree(int n) {
        return n > 0 && 1162261467 % n == 0;
    }
};

稍微改变一下，将3^0, 3^1,...,3^19这20个数存入数组中，判断n是否是数组中的某个元素即可。其实后面查找还是会用到循环或者递归。

class Solution {
public:
    bool isPowerOfThree(int n) {
        int pow3[20] = {1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907, 43046721, 129140163, 387420489, 1162261467};
        return binary_search(pow3, pow3 + 20, n);

    }
};

参考：https://leetcode.com/discuss/81699/1-line-c-no-recursion-loop，https://leetcode.com/discuss/80819/simple-solutions-without-recursion-iteration-time-and-space