技術標籤：演算法與資料結構LeetCodeleetcode位運算

Topic

Bit Manipulation

Description

https://leetcode.com/problems/power-of-four/

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: true
 

Example 2:

Input: n = 5
Output: false

Example 3:

Input: n = 1
Output: true

Constraints:

• -2³¹ <= n <= 2³¹ - 1

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

Analysis

方法1：數學除餘

方法2：位運算

對於一個整數而言，如果這個數是 4 的冪次方，那它必定也是 2 的冪次方。

從上表看出，是4的冪次方的數二進位制形式中的1在奇數位。

於是，檢測值與一個特殊的數做 & 位運算，用來判斷檢測值的二進位制形式中的1是否在奇數位。

這個特殊的數有如下特點：

• 足夠大，但不能超過 32 位，即最大為 31 個 1
• 它的二進位制表示中奇數位為 1 ，偶數位為 0

符合這兩個條件的二進位制數是1010101010101010101010101010101，轉換成0x55555555。

Submission

public class PowerOfFour {
	//方法2：
	public boolean isPowerOfFour
 
(int num) {
		return num > 0 && (num & (num - 1)) == 0 && (num & 0x55555555) != 0;
		// 0x55555555 is to get rid of those power of 2 but not power of 4
		// so that the single 1 bit always appears at the odd position
	}

	//方法1：
	public boolean isPowerOfFour2(int num) {
		while ((num != 0) && (num % 4 == 0)) {
			num /= 4;
		}
		return num == 1;
	}
}

Test

public class PowerOfFourTest {

	@Test
	public void test() {
		PowerOfFour pf = new PowerOfFour();
		
		assertTrue(pf.isPowerOfFour(4));
		assertTrue(pf.isPowerOfFour(16));
		assertFalse(pf.isPowerOfFour(17));
		assertFalse(pf.isPowerOfFour(5));
	}
	
	@Test
	public void test2() {
		PowerOfFour pf = new PowerOfFour();
		
		assertTrue(pf.isPowerOfFour2(4));
		assertTrue(pf.isPowerOfFour2(16));
		assertFalse(pf.isPowerOfFour2(17));
		assertFalse(pf.isPowerOfFour2(5));
	}
}