0693. Binary Number With Alternating Bits

693. Binary Number with Alternating Bits #

题目 #

Given a positive integer, check whether it has alternating bits: namely, if two adjacent bits will always have different values.

Example 1:

Input: 5
Output: True
Explanation:
The binary representation of 5 is: 101

Example 2:

Input: 7
Output: False
Explanation:
The binary representation of 7 is: 111.

Example 3:

Input: 11
Output: False
Explanation:
The binary representation of 11 is: 1011.

Example 4:

Input: 10
Output: True
Explanation:
The binary representation of 10 is: 1010.

题目大意 #

给定一个正整数,检查他是否为交替位二进制数:换句话说,就是他的二进制数相邻的两个位数永不相等。

解题思路 #

  • 判断一个数的二进制位相邻两个数是不相等的,即 0101 交叉间隔的,如果是,输出 true。这一题有多种做法,最简单的方法就是直接模拟。比较巧妙的方法是通过位运算,合理构造特殊数据进行位运算到达目的。010101 构造出 101010 两者相互 & 位运算以后就为 0,因为都“插空”了。

代码 #


package leetcode

// 解法一
func hasAlternatingBits(n int) bool {
	/*
	   n =         1 0 1 0 1 0 1 0
	   n >> 1      0 1 0 1 0 1 0 1
	   n ^ n>>1    1 1 1 1 1 1 1 1
	   n           1 1 1 1 1 1 1 1
	   n + 1     1 0 0 0 0 0 0 0 0
	   n & (n+1)   0 0 0 0 0 0 0 0
	*/
	n = n ^ (n >> 1)
	return (n & (n + 1)) == 0
}

// 解法二
func hasAlternatingBits1(n int) bool {
	last, current := 0, 0
	for n > 0 {
		last = n & 1
		n = n / 2
		current = n & 1
		if last == current {
			return false
		}
	}
	return true
}


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