代码如下:class Solution(object): def reorderedPowerOf2(self, N): """ :type N: int :rtype: bool """ c = collections.Counter(str(N)) return any(c == collections.Counter(str(1 << i)) for i in xrange(32)) 1 2 3 4 5 6 7
整理自leetCode 231.Power of Two && 236.Power of Three Given an integer, write a function to determine if it is a power of two. 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? 常规解题手段:...
这道题让我们判断一个数是不是3的次方数,在LeetCode中,有一道类似的题目Power of Two,那道题有个非常简单的方法,由于2的次方数实在太有特点,最高位为1,其他位均为0,所以特别容易,而3的次方数没有显著的特点,最直接的方法就是不停地除以3,看最后的迭代商是否为1,要注意考虑输入是负数和0的情况,参见代码...
解法三: classSolution {public:boolisPowerOfThree(intn) {return(n >0&∫(log10(n) / log10(3)) - log10(n) / log10(3) ==0); }}; 类似题目: Power of Two 参考资料: https://leetcode.com/discuss/78532/summary-all-solutions-new-method-included-at-15-30pm-jan-8th...
My Solutions to Leetcode problems. All solutions support C++ language, some support Java and Python. Multiple solutions will be given by most problems. Enjoy:) 我的Leetcode解答。所有的问题都支持C++语言,一部分问题支持Java语言。近乎所有问题都会提供多个算法解决。大家加油!:) 0 stars 721 forks ...
classSolution {public:boolisPowerOfFour(intnum) {returnnum >0&& !(num & (num -1)) && (num -1) %3==0; } }; 类似题目: Power of Three Power of Two 参考资料: https://leetcode.com/problems/power-of-four/ https://leetcode.com/problems/power-of-four/discuss/80457/Java-1-line-(...
powerof two power-of-two class Solution { public: bool isPowerOfTwo(int n) { return n>=1 && !(n&(n-1)); } }; n=10000***000, <=> n&(n-1)=0 是这种方法的核心 https://leetcod i++ 版本号 c++ 链表 转载 mob604756f261ee ...
power-of-two class Solution { public: bool isPowerOfTwo(int n) { return n>=1 && !(n&(n-1)); } }; n=10000***000, <=> n&(n-1)=0 是这种方法的核心 https://leetcod i++ 版本号 c++ 链表 转载 mob604756f261ee 2017-05-27 14:27:00 184阅读 2评论 Power of Four https...
500+ LeetCode Problems Solved in C++ - A repository with concise C++ solutions to over 500 LeetCode problems, perfect for preparing for coding interviews. - DSA-LeetCode/342-power-of-four/342-power-of-four.cpp at 31baa7cee7e6b0c8d68e06df7fca6238bef4274e
Leetcode 231 and 342. Power of 2, Power of 4 上原题先: 把Leetcode这两道题目放一起,其实这两个是相似的。 完全可以通过循环来做,在不能用循环的条件下如何求解? 方法一:通过考虑位运算。精妙绝伦的解法。 方法二:与方法一类似,只不过在满足num为2的幂前提下,决定num是不是4的幂的方法略微有差别...