0300 - 0399
0342 - Power of Four (Easy)
Problem Statement
Given an integer n, return true if it is a power of four. Otherwise, return false.
An integer n is a power of four, if there exists an integer x such that n == 4 ^ x.
Example 1:
Input: n = 16
Output: trueExample 2:
Input: n = 5
Output: falseExample 3:
Input: n = 1
Output: trueConstraints:
-2^31 <= n <= 2^31 - 1
Follow up: Could you solve it without loops/recursion?
Approach 1: Binary Search
class Solution {
public:
bool isPowerOfFour(int n) {
// the idea is to use binary search to find x to see if 4 ^ x = n is true or false
int l = 0, r = (int) log(pow(2, 31)) / log(4);
while (l < r) {
// get the middle one
// for even number of elements, take the lower one
int m = l + (r - l) / 2;
// exclude m
if (pow(4, m) < n) l = m + 1;
// include m
else r = m;
}
// check if 4 ^ l is n
// if so, then n is a power of four, otherwise it is not
return pow(4, l) == n;
}
};Approach 2: Bit Manipulation
class Solution {
public:
bool isPowerOfFour(int num) {
// 4: 100
// 16: 10000
// observation:
// count of 1s is 1 and the number of trailing zeros is even
return __builtin_popcount(num) == 1 && // only 1 bit is set
(__builtin_ctz(num) & 1) == 0; // with even trailing zeros
}
};