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
思路：
以 . 为分隔符分割数字，依次对比大小。注意两个版本号长度不一样的情况。
int compareVersion(string version1, string version2) {
int i = , j = ;
int n1 = , n2 = ;
while(i < version1.size() && j < version2.size()) //对比每一个小数点前对应的数字
{
n1 = , n2 = ;
while(i < version1.size() && version1[i++] != '.')
n1 = n1 * + version1[i - ] - '';
while(j < version2.size() && version2[j++] != '.')
n2 = n2 * + version2[j - ] - '';
if(n1 > n2) return ;
if(n1 < n2) return -;
}
//处理数字数量不一样多的情况 如 1.0 和 1 或 1.0.0.4 和 1.0 此时肯定比较短的那个版本号已经到头了 只要获取剩下的那个版本号后面的数字是否有大于0的即可
n1 = , n2 = ;
while(i++ < version1.size())
n1 = (version1[i - ] == '.') ? n1 : n1 * + version1[i - ] - '';
while(j++ < version2.size())
n2 = (version2[j - ] == '.') ? n2 : n2 * + version2[j - ] - '';
if(n1 > n2) return ;
else if(n1 < n2) return -;
else return ;
}
大神的代码，简洁很多。相当于把我的代码下面的循环部分和上面的融合在一起了。
public class Solution {
public int compareVersion(String version1, String version2) {
String[] v1 = version1.split("\\.");
String[] v2 = version2.split("\\.");
int longest = v1.length > v2.length? v1.length: v2.length;
for(int i=0; i<longest; i++)
{
int ver1 = i<v1.length? Integer.parseInt(v1[i]): 0;
int ver2 = i<v2.length? Integer.parseInt(v2[i]): 0;
if(ver1> ver2) return 1;
if(ver1 < ver2) return -1;
}
return 0;
}
}