Compare Version Numbers Leetcode
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Compare Version Numbers
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
Credits:
Special thanks to @ts for adding this problem and creating all test cases.
My Solution:
class Solution {
public:
int compareVersion(string version1, string version2) {
if(version1.empty() && version2.empty()) return 0;
string tmp1,tmp2;
tmp1.clear();
tmp2.clear();
int i=0,j=0;
while(version1[i]!='.' && i<version1.size()){
tmp1.push_back(version1[i++]);
}
while(version2[j]!='.' && j<version2.size()){
tmp2.push_back(version2[j++]);
}
string v1,v2;
for(int k=i+1;k<version1.size();++k)
v1.push_back(version1[k]);
for(int k=j+1;k<version2.size();++k)
v2.push_back(version2[k]);
const char *t1=tmp1.c_str();
const char *t2=tmp2.c_str();
if(atoi(t1)==atoi(t2))
return compareVersion(v1,v2);
else if(atoi(t1)<atoi(t2))
return -1;
else if(atoi(t1)>atoi(t2))
return 1;
}
};
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