[leetcode]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:

0.1 < 1.1 < 1.2 < 13.37

Credits:

Special thanks to @ts for adding this problem
and creating all test cases.

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思路是每次找到'.'然后将'.'前面的字符串转换为数字，比较如果相等就将除去前面相等数字和'.'的字符串进行递归
空串的值作为0；

前面为0的情况是要把0省略
后面为0的情况是要比较
1.0.0.0==1.0 true
01==1 true

顺便写一下字符串的几个函数吧：

string s=“12345678”;
size_t pos1=s.find('5');
//找得到返回查找字符所在位置 4
size_t pos2=s.find('0');
//找不到返回string::npos，值为INT_MAX

s.c_str()我在C++里面专门开一片文章来解释

最后注意空串和"0"的atoi都是返回数字0的

附上AC代码：

class Solution {
public:
int compareVersion(string version1, string version2) {
if(version1.empty()&&version2.empty()) return 0;
int v1,v2;
size_t pos1=version1.find('.');
size_t pos2=version2.find('.');
int ver1=atoi(version1.substr(0,pos1).c_str());
int ver2=atoi(version2.substr(0,pos2).c_str());
if(ver1>ver2)
return 1;
else if(ver1<ver2)
return -1;
else{
string nextVer1= (pos1==string::npos?"":version1.substr(pos1+1));
string nextVer2= (pos2==string::npos?"":version2.substr(pos2+1));
return compareVersion(nextVer1,nextVer2);
}
}
};