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描述
Given a positive integer num, output its complement number. The complement strategy is to flip the bits of its binary representation.
Example 1:
Input: num = 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: num = 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 num is guaranteed to fit within the range of a 32-bit signed integer.
- num >= 1
- You could assume no leading zero bit in the integer’s binary representation.
- This question is the same as 1009: leetcode.com/problems/co…
解析
根据题意，就是先找出 num 的二进制表达，然后求将其中的 1 变成 0 ，0 变成 1 形成的二进制数的十进制数。直接使用内置函数 bin 得到 num 的二进制字符串，然后遍历字符串进行上述的变化操作，最后将二进制转换成十进制返回即可。这种方法有使用内置函数的嫌疑，技术含量不高。
解答
class Solution(object):
def findComplement(self, num):
"""
:type num: int
:rtype: int
"""
s = bin(num)[2:]
r = ''
for c in s:
if c=='0':
r += '1'
else:
r += '0'
return int(r,2)
运行结果
Runtime: 20 ms, faster than 38.33% of Python online submissions for Number Complement.
Memory Usage: 13.2 MB, less than 84.58% of Python online submissions for Number Complement.
解析
因为这里需要将 0 变成 1 ，将 1 变成 0 ，最直接的我们想到了异或的操作，只要将二进制的每一位都和 1 近行异或，就能达到相同的效果。构建同样长度的二进制，用到了 << 位移操作，(1 << (len(bin(num)) - 2)) - 1 即可得到。然后将其和 num 进行 ^ 运算即可得到答案，这种方法比上面的方法要快很多。
异或运算：操作的是二进制位，两个位的数字相同为0，两个位的数字相异为1
print(1^0)
print(1^1)
print(0^0)
打印：
1
0
0
<< 左移位运算：操作的是二进制位，各二进位全部左移若干位，高位丢弃，低位补0
print(1)
print(1<<2)
print(1<<3)
打印：
1
4
8
解答
class Solution(object):
def findComplement(self, num):
"""
:type num: int
:rtype: int
"""
return num ^ ((1 << (len(bin(num)) - 2)) - 1)
运行结果
Runtime: 8 ms, faster than 98.74% of Python online submissions for Number Complement.
Memory Usage: 13.3 MB, less than 88.05% of Python online submissions for Number Complement.
原题链接：leetcode.com/problems/nu…
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