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Given an integer n, return true if it is a power of two. Otherwise, return false.

An integer n is a power of two, if there exists an integer x such that n == 2x.

Example 1:

Input: n = 1 Output: true Explanation: 2^0 = 1

Example 2:

Input: n = 16 Output: true Explanation: 2^4 = 16

Example 3:

Input: n = 3 Output: false

Example 4:

Input: n = 4 Output: true

Example 5:

Input: n = 5 Output: false

Constraints:

- -231 <= n <= 231 - 1

Follow up: Could you solve it without loops/recursion?

Title: Give you an integer n, you can judge if it is the power of 2. If it is, return true; otherwise, return false. Advanced does not use loops or recursion to complete this topic.

This question is similar to 326. Power of Three and 342. Power of Four Especially for a power like 4.

### Solution 1 Iteration+Trial

The simple solution is to keep adding n n n pairs 2 2 2 Trial until n n n is no longer associated with 2 2 2 is multiplied, then judge n n Is n equal to 2 0 = 1 2^0 = 1 20 = 1. Note that nonpositive integers are not 2 2 The time complexity of the algorithm is O ( log n ) O(\log n) O(logn), spatial complexity is O ( 1 ) O(1) O(1) :

//C++ version class Solution { public: bool isPowerOfTwo(int n) { if (n <= 0) return false; while (n % 2 == 0) n /= 2; return n == 1; } }; //Execution time: 0 ms, beating 100.00% of all C++ submissions //Memory consumption: 5.6 MB, beating 98.27% of all C++ submissions

I am here Power of 4 The sx method in can also be used here. n n n to a floating point number and divide by 2 2 2, and finally decide if it's equal to 1.0 1.0 1.0 (not recommended):

//C++ version class Solution { public: bool isPowerOfTwo(int n) { double tn = n; while (tn > 1.0) tn /= 2; return tn == 1.0; } }; //Execution time: 4 ms, beating 47.41% of all C++ submissions //Memory consumption: 5.8 MB, beating 70.78% of all C++ submissions

### Solution 2-bit operation (1 moving left continuously)

Keep going 1 1 1 Move Left 1 1 1-bit and n n n Compare until greater than or equal to n n Until n. This saves some constant time compared to using division or even floating-point division. To avoid overflow errors, use the unsigned integer type uint32_t (be careful to exclude non-positive integers, especially negative integers) or something like Power of 4 Use long long long as well. The time complexity of the algorithm is O ( log n ) O(\log n) O(logn), spatial complexity is O ( 1 ) O(1) O(1) :

//C++ version class Solution { public: bool isPowerOfTwo(int n) { if (n <= 0) return false; uint32_t x = 1; while (x < n) x <<= 1; return x == n; } }; //Execution time: 4 ms, beating 47.41% of all C++ submissions //Memory consumption: 5.8 MB, beating 66.92% of all C++ submissions

### Solution 3 Math library function (not recommended)

Use the math library functions log(), pow(), but here you need to determine if n is a positive integer and handle the error (which may be the test sample ratio) 326. Power of Three Much less, just over a thousand) and then passed:

//C++ version class Solution { public: bool isPowerOfTwo(int n) { if (n <= 0) return false; int k = log(n) / log(2); return pow(2, k) == n; } }; //Execution time: 4 ms, beating 47.41% of all C++ submissions //Memory consumption: 5.9 MB, beating 38.62% of all C++ submissions

### Solution 4 Tables

One of the easiest ways to think of "do not use loops/recursions" is to do table preprocessing. The time complexity of the algorithm is O ( 1 ) O(1) O(1), spatial complexity is O ( 1 ) O(1) O(1) :

//C++ version int ans[35] = {1,2,4,8,16,32,64,128,256,512,1024,2048,4096,8192,16384,32768,65536,131072,262144,524288,1048576,2097152,4194304,8388608,16777216,33554432,67108864,134217728,268435456,536870912,1073741824}; class Solution { public: bool isPowerOfTwo(int n) { if (n <= 0) return false; for (int i = 0; i < 32; ++i) if (n == ans[i]) return true; return false; } }; //Execution time: 0 ms, beating 100.00% of all C++ submissions //Memory consumption: 5.9 MB, beating 26.88% of all C++ submissions

### Solution 5 Mathematics (Multiple/Approximate)

n n The data type of n is int, as shown by the table, which is the largest in the int range 2 2 The second power is 1073741824 1073741824 1073741824. If n n n is 2 2 The power of 2 must be satisfied n ∗ 2 k = 1073741824 n * 2^k = 1073741824 n_2k=1073741824, that is n n n and 1073741824 1073741824 There is a multiple relationship between 1073741824. So you just need to decide n n n is a positive integer and is 1073741824 1073741824 The time complexity of the algorithm is O ( 1 ) O(1) O(1), spatial complexity is O ( 1 ) O(1) O(1) .

It is important to note that this is not a quick judgement x x General practice for the power of x if and only if x x x is a prime number, so it cannot be generalized to 342. Power of Four A simple example is, 64 64 64 is 4 4 The power of 4 must also be some of the largest 4 4 Power of 4 z z The approximate number of z, however 32 32 32 is not 4 4 The power of 4, but it must be 64 64 64, thus z z The approximate number of z.

//C++ version class Solution { public: bool isPowerOfTwo(int n) { return n > 0 && 1073741824 % n == 0; } }; //Execution time: 0 ms, beating 100.00% of all C++ submissions //Memory consumption: 5.8 MB, beating 58.76% of all C++ submissions

### Solution 6-bit operation skills

With bitwise manipulation, we can easily solve this problem. In theory, in binary form 2 2 Sum of powers of 2 4 4 Like the power of 4, there is only one binary bit 1 1 1, so you can use n > 0 to judge positive numbers, use!(n &(n - 1)) Judges that only one of the binary forms appears 1 1 1. The time complexity of the algorithm is O ( 1 ) O(1) O(1), spatial complexity is O ( 1 ) O(1) O(1) :

//C++ version class Solution { public: bool isPowerOfTwo(int n) { return n > 0 && !(n & (n - 1)); } }; //Execution time: 0 ms, beating 100.00% of all C++ submissions //Memory consumption: 5.9 MB, beating 26.88% of all C++ submissions

- Power of 2

https://leetcode-cn.com/problems/power-of-two/