[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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String

分析：将字符串中的每个数字提取出来，保存到数组中，然后逐个比较数组各个元素。

注意：“1” 和 “1.0” 以及“1.0.0” 是相同的，需要单独处理一下。。

也快参考 http://www.cnblogs.com/grandyang/p/4244123.html

class Solution {
private:
int string2Int(const string& str)
{
if(str.size() == 0)
return 0;

int rtn = 0;
for(int i = 0; i < str.size(); i++)
{
rtn = rtn*10 + (str[i] - '0');
}
return rtn;
}

vector<int> string2Ints(const string& str)
{
vector<int> rtn;
if(str.size() == 0)
return rtn;

int size = str.size();
int left = 0;

int tmp;
for(int i = 0; i < size; i++)
{
if(str[i] == '.')
{
tmp = string2Int(str.substr(left, i - left ));
rtn.push_back(tmp);
left = i + 1;
}
}
tmp = string2Int(str.substr(left, size + 1 - left ));
rtn.push_back(tmp);
return rtn;

}
public:
int compareVersion(string version1, string version2)
{
if(version1 == version2)
return false;
vector<int> rtn1 = string2Ints(version1);
vector<int> rtn2 = string2Ints(version2);

//printVector(rtn1);
//printVector(rtn2);
int size = min(rtn1.size(), rtn2.size());
for(int i = 0; i < size; i++)
{
if(rtn1[i] > rtn2[i] )
return 1;
else if(rtn1[i] < rtn2[i] )
return -1;
}
if(rtn1.size() == rtn2.size())
return 0;
else if(rtn1.size() > rtn2.size())
{
for(int i = size; i < rtn1.size(); i++)
{
if(rtn1[i] != 0)
return 1;
}
return 0;//all is zero
}
else //if(rtn1.size() < rtn2.size())
{
for(int i = size; i < rtn2.size(); i++)
{
if(rtn2[i] != 0)
return -1;
}
return 0;//all is zero
}
}
};