Compare two version numbers version1 and version1.

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  for adding this problem and creating all test cases.

给2个版本的字符串，里面只含数字和 '.' ，比较两个版本的大小。

解法：对字符串以 '.' ，进行分拆形成数组。然后从数组的第一个元素开始比较，如果相同就比较下一个元素，如果不同v1大就返回1，v2 大就返回-1，如果比较到最后都相同就返回0。

注意：如果产生的两个数组长度不一样，例如：1.2 < 1.2.3 则要把短的后面加上'0'补齐数组长度和长的一样，然后在进行比较。

Java:

String[] arr1 = version1.split("\\.");    String[] arr2 = version2.split("\\.");     int i=0;    while(i
     
       Integer.parseInt(arr2[i])){                return 1;            }        } else if(i
     

Java:

public class Solution {    public int compareVersion(String version1, String version2) {        if (version1 == null || version2 == null) return 0;        String[] v1s = version1.split("\\.");        String[] v2s = version2.split("\\.");        int i = 0, j = 0, res = 0;        while (i < v1s.length || j < v2s.length) {            int ver1 = i < v1s.length ? Integer.valueOf(v1s[i++]) : 0;            int ver2 = j < v2s.length ? Integer.valueOf(v2s[j++]) : 0;            if (ver1 < ver2) return -1;            else if (ver1 > ver2) return 1;        }        return 0;    }}

Python:

class Solution2(object):    def compareVersion(self, version1, version2):        """        :type version1: str        :type version2: str        :rtype: int        """        v1, v2 = version1.split("."), version2.split(".")                if len(v1) > len(v2):            v2 += ['0' for _ in xrange(len(v1) - len(v2))]        elif len(v1) < len(v2):            v1 += ['0' for _ in xrange(len(v2) - len(v1))]                i = 0        while i < len(v1):            if int(v1[i]) > int(v2[i]):                return 1            elif int(v1[i]) < int(v2[i]):                return -1            else:                i += 1                return 0

Python: My solution　　

class Solution(object):    """    :type v1, str    :type v2, str    :rtype: int    """    def compareVersion(self, v1, v2):        split_v1 = v1.split('.')        split_v2 = v2.split('.')        print split_v1, split_v2        n1 = len(split_v1)        n2 = len(split_v2)                if n1 < n2:            for i in xrange(n2 - n1):                split_v1 += '0'        elif n1 > n2:             for i in xrange(n1 - n2):                split_v2 += '0'                    for i in xrange(len(split_v1)):            if split_v1[i] == split_v2[i]:                continue            if split_v1[i] > split_v2[i]:                return 1            if split_v1[i] < split_v2[i]:                return -1                return 0

 

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