leetcode165. Compare Version Numbers-优快云博客

本文介绍了一种有效的版本号比较算法，通过将版本号拆分成序列并进行逐位对比，实现了快速判断两个版本号之间的大小关系。

medium程度题

题目：

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

AC解

class Solution 
{
public:
    int compareVersion(string version1, string version2) 
    {
        int i = 0;
    	long long sum1 = 0,sum2 = 0;
    	vector<long long> s1,s2;
    	
    	while(version1[i] != '\0')
    	{	//提取元素
    		if(version1[i] == '.')
    		{
    			s1.push_back(sum1);
    			sum1 = 0;
    		}
    		else
    		{
    			sum1 = 10 * sum1 + (version1[i] - '0');
    		}
    		
    		i++;
    	}
    	s1.push_back(sum1);//添加最后一个到容器
    	
    	i = 0;
    	
    	while(version2[i] != '\0')
    	{
    		if(version2[i] == '.')
    		{
    		    s2.push_back(sum2);
    			sum2 = 0;
    		}
    		else
    		{
    			sum2 = 10 * sum2 + (version2[i] - '0');
    		}
    		
    		i++;
    	}
    	s2.push_back(sum2);//添加最后一个到容器
    	
    	if(s1.size() < s2.size())
    	{
    	    s1.insert(s1.end(),s2.size() - s1.size(),0);
    	}
    	if(s1.size() > s2.size())
    	{
    	    s2.insert(s2.end(),s1.size() - s2.size(),0);
    	}
    	if(s1 == s2) return 0;
    	if(s1 > s2) return 1;
    	if(s1 < s2) return -1;
    	/*i = s1.size() < s2.size() ? s1.size():s2.size();
    	for(int j = 0 ; j < i; j++)
    	{
    	    if(s1[j] < s2[j])
    	        return -1;
    	    else if(s1[j] > s2[j])
    	        return 1;
    	}
    	
    	if(s1.size() > s2.size())
    	{
    	    for(int j = 0 ; j < fabs(s1.size() - s2.size()) ; j++)
    	    {
    	        if(s1[i] > '0')
    	            return 1;
    	    }
    	}
    	else
    	{
    	    for(int j = 0 ; j < fabs(s1.size() - s2.size()) ; j++)
    	    {
    	        if(s2[i] > '0')
    	            return -1;
    	    }
    	}
    	return 0;*/
    }
};

我的想法是把所有由.分隔的整数提取出来放入一个容器

最开始使用了注释里的解，超时！最后查文档知道容器Vector重载了一系列比较操作符，依据字典顺序比较大小

引用自http://en.cppreference.com/w/cpp/container/vector/operator_cmp

Ch if the contents of lhs and rhs are equal, that is, whether lhs.size() == rhs.size() and each element in lhs compares equal with the element in rhs at the same position.

Co compares the contents of lhs and rhs lexicographically. The comparison is performed by a function equivalent to std::lexicographical_compare.

如果容器大小相等且每个元素都相等则容器相等，具体的比较由函数 std::lexicographical_compare实现

具体细节参见 http://en.cppreference.com/w/cpp/algorithm/lexicographical_compare

为了满足题意的比较，我将容器最终大小统一，即在小的容器后面插入0使其最终与大的容器大小相同。