【LeetCode】165. Compare Version Numbers

问题描述

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

分析

看上去是只要使用split切割成数组就行了，但实际上还是有几个坑的，比如说01与1是同一个版本，1.0与1是同一个版本.

代码

class Solution {
public:
    int compareVersion(string version1, string version2) {
        vector<string> v1 = split(version1,".");
        vector<string> v2 = split(version2, ".");
        int len = max(v1.size(), v2.size()); 
        for(int i=0,j=0;i<len;i++,j++) {
        	int a = 0;
			if(i<v1.size()) {
				a = atoi(v1[i].c_str());	
			}
        	int b = 0;
			if(j<v2.size()){
				b = atoi(v2[j].c_str());
			}
        	if(a > b) {
	        	return 1;
	        } else if(a < b) {
        		return -1;
        	}
        }
        return 0;
    }
    
    std::vector<std::string> split(const  std::string& s, const std::string& delim)  
	{  
	    std::vector<std::string> elems;  
	    size_t pos = 0;  
	    size_t len = s.length();  
	    size_t delim_len = delim.length();  
	    if (delim_len == 0) return elems;  
	    while (pos < len)  
	    {  
	        int find_pos = s.find(delim, pos);  
	        if (find_pos < 0)  
	        {  
	            elems.push_back(s.substr(pos, len - pos));  
	            break;  
	        }  
	        elems.push_back(s.substr(pos, find_pos - pos));  
	        pos = find_pos + delim_len;  
	    }  
	    return elems;  
	}  
    
};