位操作运算符
C++基本的位操作符有与、或、异或、取反、左移、右移这6种,它们的运算规则如下所示:
符号 | 描述 | 运算规则 |
---|---|---|
& | 与 | 两个位都为1时,结果才为1 |
| | 或 | 两个位都为0时,结果才为0 |
^ | 异或 | 两个位相同为0,相异为1 |
~ | 取反 | 0变1,1变0 |
<< | 左移 | 各二进位全部左移若干位,高位丢弃,低位补0 |
>> | 右移 | 各二进位全部右移若干位,对无符号数,高位补0,有符号数,各编译器处理方法不一样,有的补符号位(算术右移),有的补0(逻辑右移) |
Note:
- 位操作只能用于整形数据,对float和double类型进行位操作会被编译器报错。
- 位操作符的运算优先级比较低,因为尽量使用括号来确保运算顺序,否则很可能会得到莫明其妙的结果
- 另外位操作还有一些复合操作符,如&=、|=、 ^=、<<=、>>=
位操作实战
476. Number Complement Given a positive integer, output its complement number. The complement strategy is to flip the bits of its binary representation.
Example 1:
Input: 5
Output: 2
Explanation: The binary representation of 5 is 101 (no leading zero bits), and its complement is 010. So you need to output 2.
Example 2:
Input: 1
Output: 0
Explanation: The binary representation of 1 is 1 (no leading zero bits), and its complement is 0. So you need to output 0.
Note:
- The given integer is guaranteed to fit within the range of a 32-bit signed integer.
- You could assume no leading zero bit in the integer’s binary representation.
- This question is the same as 1009: https://leetcode.com/problems/complement-of-base-10-integer/
class Solution {
public:
int findComplement(int num) {
int tmp = num;
int mask = 0;
while (tmp)
{
mask = mask << 1;
mask = mask | 1;
tmp = tmp >> 1;
}
int result = (~num) & mask;
return result;
}
};