Leetcode 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

注意版本号长度不一致的情况，虽然长度不一样，但是可能是相同的，比如1.0 == 1

class Solution {public:int compareVersion(string version1, string version2) {/*v1 v2 表示版本修改次数*/int k, temp = 0;vector<int> v1;vector<int> v2;for (int i = 0; i<version1.length(); i++){if (version1[i] == '.'){v1.push_back(temp);temp = 0;continue;}temp = temp*10 + version1[i] - '0';}v1.push_back(temp);temp = 0;for (int j = 0; j<version2.length(); j++){if (version2[j] == '.'){v2.push_back(temp);temp = 0;continue;}temp = temp*10 + version2[j] - '0';}v2.push_back(temp);for (k = 0; k < min(v1.size(), v2.size()); k++){if (v1[k]>v2[k])return 1;else if (v1[k] < v2[k])return -1;else if (v1[k] == v2[k])continue;}if (v1[k - 1] == v2[k - 1]){if (v1.size() == v2.size())return 0;else if (v1.size() > v2.size() && v1[k])return 1;else if (v1.size() > v2.size() && v1[k] == 0)return 0;else if (v1.size() < v2.size() && v2[k])return -1;else if (v1.size() < v2.size() && v2[k] == 0)return 0;}}};