[leetcode]231. Power of Two/326. Power of Three -- JavaScript 代码
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/** * @param {number} n * @return {boolean} */var isPowerOfTwo = function(n) { if(n===0){ return false; } while((n>>1)>0){ if((n & 1)==1){ return false; } n = n>>1; } if(n!=1){ return false; } return true;};
这是一道简单题,使用循环或者递归判断n是否能被2整除即可。
但是,这道题最好的方法是使用位运算,可以大大的提高代码效率。
除了位运算,第326题还提供了一种新的思路:使用Math.log函数:
/** * @param {number} n * @return {boolean} */var isPowerOfThree = function(n) { if(n===0){ return false; } var res = Math.log(n)/Math.log(3); var res_round = Math.round(res); if(Math.abs(res_round-res)<10e-15){ return true; } return false;};
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