Compare Version Numbers

最新推荐文章于 2025-08-15 18:53:39 发布

转载最新推荐文章于 2025-08-15 18:53:39 发布 · 512 阅读

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本文详细阐述了如何通过编程实现版本号之间的比较，包括获取版本号中的整数部分、处理版本号中点分隔符，并通过比较这些整数部分来得出版本号的相对大小。

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题目：leetcode

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

vector<int> getnum(const string &version)
{
	vector<int> res;
	int pre = 0;
	for (int i = 0; i < version.size(); i++)
	{
		if (version[i] == '.')
		{
			string str(version.begin() + pre, version.begin() + i);
			res.push_back(stoi(str));
			pre = i + 1;
		}
	}
	string str(version.begin() + pre, version.end());
	res.push_back(stoi(str));
	return res;
}

int compare_version(const string &version1, const string &version2)
{
	if (version1.empty() || version2.empty())
		throw exception();

	int count1 = 0, count2 = 0;
	for (int i = 0; i<version1.size(); i++)
	{
		if (version1[i] == '.')
			count1++;
	}
	for (int i = 0; i<version2.size(); i++)
	{
		if (version2[i] == '.')
			count2++;
	}
	if (count1>count2)
		return -1 * compare_version(version2, version1);

	vector<int> num1 = getnum(version1);
	vector<int> num2 = getnum(version2);
	int index = 0;
	while (index < num1.size())
	{
		if (num1[index] == num2[index])
			index++;
		else
			break;
	}
	if (index == num1.size())
	{
		if (num1.size() == num2.size())
			return 0;
		//注意version2后面全是零的情况！！
		while (index < num2.size())
		{
			if (num2[index] != 0)
				return -1;
			index++;
		}
		//注意version2后面全是零的情况！！如 1.2 与 1.2.0.0.0
		return 0;
	}
	else
	{
		if (num1[index] > num2[index])
			return 1;
		else if (num1[index] < num2[index])
			return -1;
		else
			return 0;
	}

}