[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。