LeetCode-Compare Version Numbers

Compare Version Numbers

##题目

####Compare Version Numbers

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

####Credits:
Special thanks to @ts for adding this problem and creating all test cases.

##解题思路
该题主要是字符串的操作，我们从主版本号到次版本号依次进行比较。但是此时需要考虑一个特殊情况，即‘1.0’和‘1’这种情况，在本题中，这种情况认为是相等。

##算法代码
代码采用JAVA实现：

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public class Solution {
    public int compareVersion(String version1, String version2) {
    	//小数点在字符串分割中是特殊符号，需要在前面加"\\"
        String[] ver1=version1.split("\\.");
        String[] ver2=version2.split("\\.");
        int i=0;
        int j=0;
        //从主版本号到次版本号依次进行比较
        while(i<ver1.length && j<ver2.length)
        {
        	if(Integer.parseInt(ver1[i])<Integer.parseInt(ver2[j]))
        		return -1;
        	else if(Integer.parseInt(ver1[i])>Integer.parseInt(ver2[j]))
        		return 1;
        	else
        	{
        		i++;
        		j++;
        	}
        }
        //version1比较长
        if(i<ver1.length)
        {
        	//特殊处理‘1.0’和‘1’这种情况
        	if(Integer.parseInt(ver1[i])==0)
        		return 0;
        	else
        		return 1;
        }
        //version2比较长
        if(j<ver2.length)
        {
        	//特殊处理‘1.0’和‘1’这种情况
        	if(Integer.parseInt(ver2[j])==0)
        		return 0;
        	else
        		return -1;
        }
        return 0;        
    }
}

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