Compare Version Numbers- LeetCode 165

最新推荐文章于 2018-09-18 01:37:00 发布

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最新推荐文章于 2018-09-18 01:37:00 发布

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分类专栏： OJ 文章标签： leetcode algorithm

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本文介绍了一种用于比较软件版本号大小的方法，通过将版本号分解为数字序列并逐一比较来确定其相对大小。提供了C++和Java两种实现方式，并对比了它们的运行效率。

题目描述：

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：该题是纯粹的字符串处理题目，要求比较两个字符串表示的版本号的大小。版本号是将多个字符数字用'.'号连起来的字符串。

比较方法：用'.'号将版本号分割成多个数字字符串，然后对应下标逐一比较即可。

需注意：

1、处理每个子串首尾的0，

2、若用用java实现，将字符串按"."分割的正则表达式为"\\."，而不是"."。

3、若用C++ 实现，需自己写字符串分割函数，但是比起java，运行时间简直快太多了。

以下是C++和java实现代码的对比，附运行时间：

/* C++实现：运行耗时3ms*/

<span style="font-size:14px;">#include<iostream>
#include<string>
#include<vector>
#include<string>
using namespace std;

class Solution {
public:
        void output(vector<string> v)  /* 输出一个向量v */
	{
		for(vector<string>::iterator itr = v.begin(); itr != v.end(); itr++)
			cout << *itr << " ";
		cout << endl;
	}
	int s2i(string v) /* 将数字字符串转化成整数，忽略前端的全部0  */
	{
		int num = 0;
		for(string::size_type ix = 0; ix != v.size(); ix++)
		{
			num = num * 10 + (v[ix]-'0');
		}
	//	cout << v << ":" <<  num << endl;
		return num;
	}

	int compare(string v1,string v2) 
	{
		int num1 = s2i(v1);
		int num2 = s2i(v2);
		if(num1 > num2)
			return 1;
		else if(num1 < num2)
			return -1;
		else 
			return 0;
	}

        /* 用字符char分割版本号，返回一个字符串向量 */
	vector<string> strSplit(string version1,char ch) 
	{
		vector<string> strv;
		string v = version1;
		int pos = 0;
		while((pos = v.find_first_of(ch,0) ) != string::npos)
		{
			string s = v.substr(0,pos);
			v = v.substr(pos+1);
			strv.push_back(s);
		}
		strv.push_back(v);
		//output(strv);
		return strv;
	}
	
        /* 调用辅助函数，比较两个版本号的大小 */
	int compareVersion(string version1, string version2) 
	{	
		int flag = 0;
		vector<string> v1 = strSplit(version1, '.');
		vector<string> v2 = strSplit(version2, '.');
	
		vector<string>::size_type len1 = v1.size();
	        vector<string>::size_type len2 = v2.size();
		if(len1 > len2)
		{
			vector<string>::size_type ix = 0;
			for( ix = 0; ix != len2; ix++)
			{
				flag = compare(v1[ix] , v2[ix]);
				if(flag != 0)
					return flag;
			}
			while(ix < len1)
			{
				if(s2i(v1[ix]) != 0)
				{
					flag = 1;
					return flag;
				}
				ix++;
			}
		}
		else
		{
			vector<string>::size_type ix = 0;
			for( ix = 0; ix != len1; ix++)
			{
				flag = compare(v1[ix] , v2[ix]);
				if(flag != 0)
					return flag;
			}
			while(ix < len2)
			{
				if(s2i(v2[ix]) != 0)
				{
					flag = -1;
					return flag;
				}
				ix++;
			}	
		}
		return flag;
	}
};
int main ()
{
	Solution solution;
	string v1 = "1.01.0";
	string v2 = "1.1";
	
	cout <<solution.compareVersion(v1,v2);
	system("pause");
	return 0;
}</span>

/* Java 实现：运行耗时287ms */

<span style="font-size:14px;">package test;
public class Solution {
	public static void main(String[] args) {
		String v1 = "1.1";
		String v2 = "1.01.0";
		System.out.println(compareVersion(v1, v2) );
	}
	public static int s2i(String v)
	{
		int num = 0;
		for(int ix = 0; ix != v.length(); ix++)
		{
			num = num * 10 + (v.charAt(ix)-'0');
		}
		return num;
	}
	public static int compare(String v1,String v2)
	{
		int num1 = s2i(v1);
		int num2 = s2i(v2);
		if(num1 > num2)
			return 1;
		else if(num1 < num2)
			return -1;
		else 
			return 0;
	}
	public static int compareVersion(String version1, String version2) {
		int flag = 0;
		String[] v1 = version1.split("\\.");
		String[] v2=  version2.split("\\.");
		int len1 = v1.length;
		int len2 = v2.length;	
	//	System.out.println(len1 + " "  + len2);
		if(len1 -len2 < 0)
		{
			int i =0;
			for(i = 0; i < len1 ; i++)
			{
				flag = compare(v1[i] , v2[i]);
				if( flag != 0 )
				{
					return flag;
					//break;
				}
			}
			while(i < len2)
			{
				if(s2i(v2[i]) != 0)
				{
					flag = -1;
					//break;
					return flag;
				}
				i ++;
			}
		}
		else
		{
			int i = 0;
			for( i= 0; i < len2 ; i++)
			{
				flag = compare(v1[i] , v2[i]);
				if( flag != 0 )
				{
					//break;
					return flag;
				}
			}
			while(i < len1)
			{
				if(s2i(v1[i]) != 0)
				{
					flag = 1;
					//break;
					return flag;
				}
				i ++;
			}	
		}
		return flag;
	}
}</span>