Leetcode Compare version numbers

最新推荐文章于 2016-12-18 10:34:10 发布

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本文介绍了一种用于比较两个版本号大小的方法。通过解析版本号字符串并逐段对比数字来判断版本1与版本2之间的大小关系，返回相应的比较结果。文章提供了一个Java实现的例子。

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

注意一下 .数目不一样的时候用0代替就好
[code]

public class Solution {
    public int compareVersion(String version1, String version2) 
    {
        int i=0, j=0;
        while(i<version1.length() || j<version2.length())
        {
            int v1=0, v2=0;
            if(i==version1.length())v1=0;
            else 
            {
                int temp=i;
                while(i<version1.length() && version1.charAt(i)!='.')i++;
                v1=Integer.parseInt(version1.substring(temp,i));
            }
            if(i<version1.length())i++;
            if(j==version2.length())v2=0;
            else
            {
                int temp=j;
                while(j<version2.length() && version2.charAt(j)!='.')j++;
                v2=Integer.parseInt(version2.substring(temp,j));
            }
            if(j<version2.length())j++;
            if(v1<v2)return -1;
            else if(v1>v2)return 1;
        }
        return 0;
    }

    boolean valid(String s)
    {
        return true;

    }
}