[leetcode] 231. Power of Two
2016-03-20 21:25
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Given an integer, write a function to determine if it is a power of two.
这道题是判断数字是否是2的乘方,题目难度为Easy。
int范围内2的乘方只有31个,逐个判断即可,具体代码:
另外,假定N为2的乘方,则N&(N-1)为0,也可以依此判断数字是否是N的乘方,具体代码:
这道题是判断数字是否是2的乘方,题目难度为Easy。
int范围内2的乘方只有31个,逐个判断即可,具体代码:
class Solution { public: bool isPowerOfTwo(int n) { for(int i=0; i<31; ++i) { if(n == 1<<i) return true; } return false; } };
另外,假定N为2的乘方,则N&(N-1)为0,也可以依此判断数字是否是N的乘方,具体代码:
class Solution { public: bool isPowerOfTwo(int n) { if(n <= 0) return false; return !(n&(n-1)); } };
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