小菜鸡终于抓住了一道会写的题。然而也是遇到了不少问题。
首先，将leetcode上的回答格式粘贴到sublime里会变成空格和tab混用的状态，于是各种报indent error。这个最后用sublime里view->indentation->convert indentation to spaces解决了。
然后的代码是（错误代码错误代码错误代码）：
class Solution(object): def hammingDistance(self, x, y): if x>y: x,y=y,x xb = bin(x).lstrip('0b') yb = bin(y).lstrip('0b') count = int(0) for i in range(1,len(yb)+1): if i<=len(xb): if xb[-i]!=yb[-i]: count+=1 elif yb[-i]==1: count+=1 return count各种错误。比如输入0，6，输出应为2。然而结果为0。很是苦恼。思索了半天，发现问题在elif一句中，yb是str类，用来比较的应该是str的1，即'1'。尴尬。后更改，ac。
class Solution(object): def hammingDistance(self, x, y): if x>y: x,y=y,x xb = bin(x).lstrip('0b') yb = bin(y).lstrip('0b') count = int(0) for i in range(1,len(yb)+1): if i<=len(xb): if xb[-i]!=yb[-i]: count+=1 elif yb[-i]=='1': count+=1 return count这一道题也算是让我一下体验了python这门语言的两个大坑了——indent和弱类型。
下一步看看discuss里大神们是怎么实现的。
看了discuss，果然是被虐了。自己想到bin方法的确不错，但是没有想到python的位运算符们：
参考yuyuyu0905的答案，使用按位异或运算符结合str.count()方法，即可得到一行python的答案：
def hammingDistance(self, x, y): """ :type x: int :type y: int :rtype: int """ return bin(x^y).count('1')另外还看到了YDD9的答案：list comprehension in python is fast
return len([i for i in format(x ^ y, 'b') if i=='1'])其中 format 是built-in方法：
format ( value [, format_spec ] )Convert a value to a “formatted” representation, as controlled by format_spec. The interpretation of format_spec will depend on the type of the value argument, however there is a standard formatting syntax that is used by most built-in types: Format Specification Mini-Language.
The default format_spec is an empty string which usually gives the same effect as calling str(value).
A call to format(value, format_spec) is translated to type(value).__format__(value, format_spec) which bypasses the instance dictionary when searching for the value’s __format__() method. A TypeError exception is raised if the method search reachesobject and the format_spec is non-empty, or if either the format_spec or the return value are not strings.
列表生成式后面也可以加上if判断。