【leetcode】476 Number Complement

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.

搞了半天位运算，从小到大都搞不清位运算，专开一章分析吧 最后还是字串弄的

93%

note：replace()函数需要赋值，不是直接在字串上操作

note：二进制字串转十进制利用int() 

十进制转二进制字串利用bin()  #注意开头是0b

class Solution(object):
    def findComplement(self, num):
        """
        :type num: int
        :rtype: int
        """
        string = bin(num)[2:]
        string = string.replace('1','a')
        string = string.replace('0','1')
        string = string.replace('a','0')
        return int(string,2)

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果然，利用亦或运算

比如101，则用111^101，结果为010

问题只剩下构建111，利用左移位运算，最后再减1，最后即可得到所有都为1的二进制数

不过效果没有直接在字串上操作那么好 

80.25%

class Solution(object):
    def findComplement(self, num):
        """
        :type num: int
        :rtype: int
        """
        i = 1
        while i<=num:
            i = i<<1
        return (i-1)^num

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或者直接用bit_length()函数   42.58%

class Solution(object):
    def findComplement(self, num):
        """
        :type num: int
        :rtype: int
        """
        i = 1
        i = i<<(num.bit_length())
        return (i-1)^num

----------------------------------------

97.2%    说明还是自己len()函数测量比较高效

class Solution(object):
    def findComplement(self, num):
        """
        :type num: int
        :rtype: int
        """
        i = 1
        i = i<<(len(bin(num))-2)
        return (i-1)^num

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1.54% 吓一跳 这种把位运算弄成字串转int的写法很耗时

class Solution(object):
    def findComplement(self, num):
        """
        :type num: int
        :rtype: int
        """
        i = '1'*len(bin(num)[2:])
        return int(i,2)^num

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format这个函数了不得，这里可以用来进制转换

eg:   print "int: {0:d}; hex: {0:x}; oct: {0:o}; bin: {0:b}".format(42)

结果：'int: 42; hex: 2a; oct: 52; bin: 101010'

print "int: {0:d}; hex: {0:#x}; oct: {0:#o}; bin: {0:#b}".format(42)

结果：'int: 42; hex: 0x2a; oct: 0o52; bin: 0b101010'

其中o是八进制，x是十六进制，b是二进制，啥符号没有是十进制

68.35%

class Solution(object):
    def findComplement(self, num):
        """
        :type num: int
        :rtype: int
        """
        return int(''.join(['1' if i=='0' else '0' for i in '{0:b}'.format(num)]),2)
        #for i in '{0:b}'.format(num):
         #   if i=='0':
          #      result.append('1')
           # else:
            #    result.append('0')
            #num = ''.join(result)
        #return int(num,2)