LeetCode 693 Binary Number with Alternating Bits 带有交替位的二进制数
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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.
题目解析
对于给定的正整数,判断其是否有交替的二进制位,即:是否相邻的位有不同的值
解题思路
最简单的方法就是一位一位的进行判断,用变量bit记录上一个位置的值,初始值为-1,与1做与运算得到最低位的值,判断相邻两个值是否相同,相同返回false,循环退出后返回true。
/* C++ */class Solution {public: bool hasAlternatingBits(int n) { int bit = -1; while (n > 0) { if (n & 1 == 1) { if (bit == 1) return false; bit = 1; } else { if (bit == 0) return false; bit = 0; } n >>= 1; } return true; }};
另一种解法,利用0和1的交替特性,进行错位相加,组成全1的二进制数,然后再用一个检测全1的二进制数的trick,就是‘与’上加1后的数,因为全1的二进制数加1,就会进一位,并且除了最高位,其余位都是0,跟原数相‘与’就会得0,所以我们可以这样判断。比如n是10101,那么n>>1就是1010,二者相加就是11111,再加1就是100000,二者相‘与’就是0
/* C++ */class Solution {public: bool hasAlternatingBits(int n) { return ((n + (n >> 1) + 1) & (n + (n >> 1))) == 0; }};
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