leetcode 165

165 easy Compare Version Numbers

*Compare two version numbers version1 and version2.

If version1 > version2 return 1, if version1 < version2 return -1, otherwise return 0.

You may assume that the version strings are non-empty and contain only digits and the . character.

The . character does not represent a decimal point and is used to separate number sequences.

For instance, 2.5 is not “two and a half” or “half way to version three”, it is the fifth second-level revision of the second first-level revision.

Here is an example of version numbers ordering:*

class Solution(object):
def compareVersion(self, version1, version2):
"""
:type version1: str
:type version2: str
:rtype: int
"""
l1=version1.split('.')
l2=version2.split('.')
n1=len(l1)
n2=len(l2)
i=0
while i<n1 and i<n2:
if int(l1[i])<int(l2[i]):
return -1
elif int(l1[i])>int(l2[i]):
return 1
else:
i+=1
if n1==n2:
return 0
elif n1<n2:
return -1 if len([v for v in l2[i:] if int(v)!=0])>0 else 0
elif n1>n2:
return 1 if len([v for v in l1[i:] if int(v)!=0])>0 else 0

这道题刚开始自己写的时候 没有考虑会有两个比较的数用split分开后个数不同的情况 在参考了别人的程序之后 发现时会有这种情况发生的 所以要有两个变量来存储这两个要比较的数的长度 还有就是要注意 split()函数 返回的是字符串数组 要比较两个数的大小时要用int来抢转换 不然的话 比如n1=’01.1’ n2=’1.1’则会出现错误。