Compare two version numbers version1 and version1.
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
class Solution {
public:
string deletezero(string str)
{
if(str.length() < 2) return str;
int i = 0;
while(i < str.length() && str[i] == '0')
i++;
if(i == str.length())
return "0";
return string(str, i);
}
int compareVersion(string version1, string version2) {
if(version1 == version2) return 0;
int dot1 = 0, dot2 = 0;
while(dot1 < version1.size() && version1[dot1] != '.')
dot1++;
while(dot2 < version2.size() && version2[dot2] != '.')
dot2++;
string sv1 = deletezero(string(version1, 0, dot1));
string sv2 = deletezero(string(version2, 0, dot2));
if(sv1.length() != sv2.length())
return sv1.length() > sv2.length() ? 1 : -1;
else if(sv1 != sv2)
return sv1 > sv2 ? 1 : -1;
else{//sv1 == sv2, cpmpare what behaind dot
string zerodot = (".0");
if(dot1 == version1.length() && dot2 == version2.length())
return 0;
else if(dot1 != version1.length() && dot2 == version2.length()){
string tmp = string(version1, dot1+1);
if(tmp.find_first_not_of(zerodot) == string::npos)return 0;
else return 1;
}else if(dot1 == version1.length() && dot2 != version2.length()){
string tmp = string(version2, dot2+1);
if(tmp.find_first_not_of(zerodot) == string::npos)return 0;
else return -1;
}
sv1 = string(version1, dot1+1);
sv2 = string(version2, dot2+1);
return compareVersion(sv1, sv2);
}
}
};