leetcode 165. Compare Version Numbers

我不是女神ヾ 2022-07-27 11:40 146阅读 0赞

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 {
    	int my_atoi(string s)
    	{
    		int r = 0;
    		while (!s.empty())
    		{
    			r = 10 * r + s[0] - '0';
    			s.erase(0, 1);
    		}
    		return r;
    	}
    public:
    	int compareVersion(string version1, string version2) {
    		if (version1 == version2)
    			return 0;
    		int pos1 = version1.find('.');
    		if (pos1 == string::npos)
    			pos1 = version1.length();
    		int pos2 = version2.find('.');
    		if (pos2 == string::npos)
    			pos2 = version2.length();
    		
    		int r1 = my_atoi(string(version1.begin(), version1.begin() + pos1));
    		int r2 = my_atoi(string(version2.begin(), version2.begin() + pos2));
    		if (r1 > r2)
    			return 1;
    		if (r2 > r1)
    			return -1;
    		
    		version1.erase(0, pos1 + 1);
    		version2.erase(0, pos2 + 1);
    		if (version1.empty() && version2.empty())
    			return 0;
    		if (version1.empty())
    		{
    			for (int i = 0; i < version2.length(); i++)
    				if (version2[i] != '.'&&version2[i] != '0')
    					return -1;
    			return 0;
    		}
    		if (version2.empty())
    		{
    			for (int i = 0; i < version1.length(); i++)
    				if (version1[i] != '.'&&version1[i] != '0')
    					return 1;
    			return 0;
    		}
    		return compareVersion(version1,version2);
    	}
    };