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80 lines
1.7 KiB
Markdown
Executable File
80 lines
1.7 KiB
Markdown
Executable File
# [693. Binary Number with Alternating Bits](https://leetcode.com/problems/binary-number-with-alternating-bits/)
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## 题目
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Given a positive integer, check whether it has alternating bits: namely, if two adjacent bits will always have different values.
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**Example 1**:
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Input: 5
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Output: True
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Explanation:
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The binary representation of 5 is: 101
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**Example 2**:
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Input: 7
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Output: False
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Explanation:
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The binary representation of 7 is: 111.
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**Example 3**:
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Input: 11
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Output: False
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Explanation:
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The binary representation of 11 is: 1011.
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**Example 4**:
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Input: 10
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Output: True
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Explanation:
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The binary representation of 10 is: 1010.
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## 题目大意
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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。
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## 解题思路
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- 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。
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## 代码
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```go
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package leetcode
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// 解法一
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func hasAlternatingBits(n int) bool {
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/*
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n = 1 0 1 0 1 0 1 0
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n >> 1 0 1 0 1 0 1 0 1
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n ^ n>>1 1 1 1 1 1 1 1 1
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n 1 1 1 1 1 1 1 1
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n + 1 1 0 0 0 0 0 0 0 0
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n & (n+1) 0 0 0 0 0 0 0 0
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*/
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n = n ^ (n >> 1)
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return (n & (n + 1)) == 0
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}
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// 解法二
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func hasAlternatingBits1(n int) bool {
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last, current := 0, 0
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for n > 0 {
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last = n & 1
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n = n / 2
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current = n & 1
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if last == current {
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return false
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}
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}
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return true
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}
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``` |