[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

解题思路：
首先求用‘.’分隔的版本号的加权和，随后比较加权和的大小。

实现代码：

/*****************************************************************************
*  @COPYRIGHT NOTICE
*  @Copyright (c) 2015, 楚兴
*  @All rights reserved
*  @Version  : 1.0

*  @Author   : 楚兴
*  @Date     : 2015/2/6 21:52
*  @Status   : Accepted
*  @Runtime  : 4 ms
*****************************************************************************/
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
class Solution {
public:
int compareVersion(string version1, string version2) {
double v1 = strtod(version1);
double v2 = strtod(version2);
if (v1 == v2)
{
return 0;
}
return v1 > v2 ? 1 : -1;
}

double strtod(string str)
{
int i = 0;
double sum = 0;
double ratio = 1;
int num = 0;
while (i < str.size())
{
if (str[i] == '.')
{
sum += num / ratio;
num = 0;
ratio *= 10; //权重降级
}

if (str[i] >= '0' && str[i] <= '9')
{
num = num * 10 + str[i] - '0';
}
i++;
}
sum += num / ratio;
return sum;
}
};

int main()
{
Solution s;
cout<<s.compareVersion("1","0")<<endl;
cout<<s.compareVersion("0.1","1.1")<<endl;
cout<<s.compareVersion("1.1","1.2")<<endl;
cout<<s.compareVersion("13.37","9.4")<<endl;
cout<<s.compareVersion("1.1","1.15")<<endl;
cout<<s.compareVersion("1.1","1.01.0")<<endl;
system("pause");
}

上述程序虽然能够被LeetCode接受，但实际上还是存在问题。在降权的时候，权重问题不好确定。目前的程序在处理版本号“1.2000”和"2.1"时会给出错误结果。同时如果版本号级数较多，会存在小数部分存储不完的情况。故更改如下：class Solution {
public:
int compareVersion(string version1, string version2) {
int i = 0;
int j = 0;
while (i < version1.size() || j < version2.size())
{
int num1 = 0;
while(i < version1.size())
{
if (version1[i] == '.')
{
i++;
break;
}
else
{
num1 = num1 * 10 + version1[i] - '0';
i++;
}
}

int num2 = 0;
while(j < version2.size())
{
if (version2[j] == '.')
{
j++;
break;
}
else
{
num2 = num2 * 10 + version2[j] - '0';
j++;
}
}

if (num1 != num2)
{
return num1 > num2 ? 1 : -1;
}
}
return 0;
}
};
Runtime: 3 ms

分别对比每一段的版本号，如果相同则继续比较下一段。不同则返回。