Problem Definition:
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:
0.1 < 1.1 < 1.2 < 13.37
Solution:
1 def compareVersion(self,version1, version2): 2 f1=map(lambda x: int(x) ,version1.split('.')) 3 f2=map(lambda x: int(x) ,version2.split('.')) 4 f1+=([0]*(len(f2)-len(f1))) 5 f2+=([0]*(len(f1)-len(f2))) 6 return 1 if f1>f2 else (-1 if f1<f2 else 0)
Tricks:
1) b=[0]*a，当a是大于0的整数时，b是一个有a个0的数组；当a<=0，b==[ ]
2) a=[1,2,3] b=[4,5,6] a+b==[1,2,3,4,5,6] 相当于a.extend(b)
3) a=[1,4] b=[2,3] 则a<b
a=[1,4] b=[1,3] 则a>b
a=[1] b=[0,2] 则a>b
a=[1] b=[1,0] 则a<b
从左边其，从俩数组a,b中分别取同下标的元素ai，bi；比较其大小，若ai>bi，则a>b；若ai，bi，则a<b；否则下标加1，继续下一次比较...
如果到了一个数组结束仍未比出大小，则判定先结束的数组小于后结束的。
4）三目运算可以这么嵌套 a=b if c else ( d if e else f）
...应该算是...干货吧...