Leetcode 165. Compare Version Numbers (Easy) (cpp)

最新推荐文章于 2024-06-04 07:48:41 发布

Niko_Ke

最新推荐文章于 2024-06-04 07:48:41 发布

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分类专栏： Leetcode C++ C++ Leetcode String 文章标签： leetcode cpp

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本文介绍了一种使用C++实现的LeetCode第165题版本号比较的方法。该方法通过将版本号字符串拆分为整数序列，并逐个比较这些整数来确定两个版本号之间的大小关系。

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Leetcode 165. Compare Version Numbers (Easy) (cpp)

Tag: String

Difficulty: Easy

/*

165. Compare Version Numbers (Easy)

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

*/
class Solution {
public:
	int compareVersion(string version1, string version2) {
		vector<int> ver1 = split(version1, '.');
		vector<int> ver2 = split(version2, '.');
		int size_min = min(ver1.size(), ver2.size()), index = 0;
		for (index; index < size_min; index++) {
			if (ver1[index] > ver2[index])
				return 1;
			if (ver1[index] < ver2[index])
				return -1;
		}
		if (ver1.size() > ver2.size()) {
			while(!ver1[index] && index < ver1.size())
				index++;
			if (index < ver1.size())
				return 1;
		}
		if (ver1.size() < ver2.size()) {
			while(!ver2[index] && index < ver2.size())
				index++;
			if (index < ver2.size())
				return -1;
		}
		return 0;
	}
	vector<int> split(string s, char sign) {
		vector<int> res;
		stringstream str(s);
		string item;
		while(getline(str, item, sign)) {
			res.push_back(atoi(item.c_str()));
		}
		return res;
	}
};