QUESTION
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
1ST TRY
class Solution {
public:
int compareVersion(string version1, string version2) {
int integer1 = ;
int integer2 = ;
int i1 = ;
int i2 = ; while(i1 < version1.length())
{
integer1 = *integer1 + version1[i1++] - '';
if(version1[i1] == '.')
{
i1++;
break;
}
}
while(i2 < version2.length())
{
integer2 = *integer2 + version2[i2++] - '';
if(version2[i2] == '.')
{
i2++;
break;
}
}
if(integer1 > integer2) return ;
else if(integer1 < integer2) return -; integer1 = ;
integer2 = ;
while(i1 < version1.length())
{
integer1 = *integer1 + version1[i1++] - '';
}
while(i2 < version2.length())
{
integer2 = *integer2 + version2[i2++] - '';
}
if(integer1 > integer2) return ;
else if(integer1 < integer2) return -;
else return ;
}
};
Result: Wrong
Input: "1.1", "1.01.0"
Output: -1
Expected: 0
2ND TRY
考虑有不只一个小数点的情况
class Solution {
public:
int compareVersion(string version1, string version2) {
int integer1 = ;
int integer2 = ;
int i1 = ;
int i2 = ; while(i1 < version1.length() || i2 < version2.length())
{
while(i1 < version1.length())
{
integer1 = *integer1 + version1[i1++] - '';
if(version1[i1] == '.')
{
i1++;
break;
}
}
while(i2 < version2.length())
{
integer2 = *integer2 + version2[i2++] - '';
if(version2[i2] == '.')
{
i2++;
break;
}
}
if(integer1 > integer2) return ;
else if(integer1 < integer2) return -; integer1 = ;
integer2 = ;
}
return ;
}
};
Result: Accepted