[LeetCode] 165. Compare Version Numbers

Compare Version Numbers

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

Solution:

 split字符串生成list，长度分别为len1, len2，在最小长度范围内比较对应的数值，记得将list元素转化为int

 之后（此时最小长度内各元素均相等）在minLen ~ maxLen范围内检测是否有非0元素，如有，则继续判断1/-1，否则返回0.

Running Time: O(n)

class Solution:
    # @param version1, a string
    # @param version2, a string
    # @return an integer
    def compareVersion(self, version1, version2):
        v1 = version1.split('.')
        v2 = version2.split('.')
        minLen = min(len(v1), len(v2))
        maxLen = max(len(v1), len(v2))
        for i in range(minLen):
            if int(v1[i]) > int(v2[i]):
                return 1
            if int(v1[i]) < int(v2[i]):
                return -1
        for i in range(minLen, maxLen):
            if len(v1) > len(v2) and int(v1[i]) != 0:
                return 1
            if len(v2) > len(v1) and int(v2[i]) != 0:
                return -1
        return 0