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@ChristopherWu 2016-06-16T21:22:28.000000Z 字数 1810 阅读 1381

Power of Two

leetcode


231. Power of Two

Question

Given an integer, write a function to determine if it is a power of two.

Solution

Approach #1 (count the number of 1) [Accepted]

Algorithm

Integer is a power of two means only one bit of n is '1', for example, 100 is 2^2=4 while 110 is 2^2+2^1=6.

When n<=0, it can't be power of two as 2^-1=0.5 and because the parameter n is int, we can sure that it has only 32 bit, so we count the number of 1 in 32 bits to check whether only one bit of n is '1' when n is positive.

  1. class Solution {
  2. public:
  3. bool isPowerOfTwo(int n) {
  4. if(n<=0)
  5. return false;
  6. int nums_of_one = 0;
  7. for(int i=0; i<32; ++i) {
  8. nums_of_one += n&1;
  9. if(nums_of_one > 1)
  10. return false;
  11. n >>= 1;
  12. }
  13. return true;
  14. }
  15. };

Complexity Analysis


Approach #2 (log2 in C++11) [Accepted]

Algorithm

The result of log2(n) in math must be an interger instead of float when integer is a power of two, so we use log2() function in C++11 and check whether log2(n) is an interger by difference between floor(log2(n)) and ceil(log2(n)).

For example, n=5, log2(5)=2.19722, floor(2.19722)=2, ceil(2.19722)=3, the difference is 1, so it is not power of two.

  1. class Solution {
  2. public:
  3. bool isPowerOfTwo(int n) {
  4. if(n<=0)
  5. return false;
  6. double tmp = log2(n);
  7. int a = floor(tmp);
  8. int b = ceil(tmp);
  9. if(b-a == 0)
  10. return true;
  11. return false;
  12. }
  13. };

Complexity Analysis

We can also use pow(2, log2(n)) instead of floor(log2(n)) - ceil(log2(n)).

  1. class Solution {
  2. public:
  3. bool isPowerOfTwo(int n) {
  4. if(!n)
  5. return false;
  6. int t = floor(log2(n));
  7. if(pow(2, t) == n)
  8. return true;
  9. return false;
  10. }
  11. };

Approach #3 (using n&(n-1) trick) [Accepted]

Algorithm

I didn't come up with this, thanks for dong.wang.1694's solution.

We can know that power of 2 means only one bit of n is '1', for example, 1, 10, 100 etc, so n-1 means the other bits will become 1, e.g. 0, 01, 011. Power of 2 minus 1 means all of its digits will negate.
Therefore, the result of n&(n-1) must be 0.

  1. class Solution {
  2. public:
  3. bool isPowerOfTwo(int n) {
  4. if(n<=0) return false;
  5. return !(n&(n-1));
  6. }
  7. };

Complexity Analysis

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