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
这个题目的难点在于其各种特殊情况，如000.0和0,返回值是0,001和1返回值也是0,10.1和10.1.0返回值也是0
我解决的办法是找到.这个分界点如果比出来了就可以得到返回值，如果前面的值的大小是相同的，则递归调用，传递的参数是去除前面的相等的数据之后的字符串
class Solution { public: int compareVersion(string version1, string version2) { size_t pos1 = 0, pos2 = 0; int num1 = 0, num2 = 0; if ((count(version1.cbegin(), version1.cend(), ‘0‘) + count(version1.cbegin(), version1.cend(), ‘.‘)) == version1.length()) { if (count(version2.cbegin(), version2.cend(), ‘0‘) + count(version2.cbegin(), version2.cend(), ‘.‘) == version2.length()) return 0; else return -1; } else { if (count(version2.cbegin(), version2.cend(), ‘0‘) + count(version2.cbegin(), version2.cend(), ‘.‘) == version2.length()) return 1; else { pos1 = version1.find_first_not_of(‘0‘); pos2 = version2.find_first_not_of(‘0‘); version1 = version1.substr(pos1); version2 = version2.substr(pos2); if (version1 == version2)return 0; pos1 = version1.find_first_of(‘.‘); pos2 = version2.find_first_of(‘.‘); if (pos1 != 0&&pos1!=string::npos) num1 = stoi(version1.substr(0,pos1)); else if (pos1 == string::npos) { num1 = stoi(version1.substr(0)); } else num1 = 0; if(pos2!=0&&pos2!=string::npos) num2 = stoi(version2.substr(0,pos2)); else if (pos2 == string::npos) { num2 = stoi(version2.substr(0)); } else num2 = 0; if (num1 == num2&&pos1 == string::npos&&pos2 == string::npos)return 0; if (num1 == num2&&pos1 == string::npos) { version2 = version2.substr(pos2); if ((count(version2.cbegin(), version2.cend(), ‘.‘) + count(version2.cbegin(), version2.cend(), ‘0‘)) == version2.length()) return 0; else return -1; } if (num1 == num2&&pos2 == string::npos) { version1 = version1.substr(pos1); if ((count(version1.cbegin(), version1.cend(), ‘.‘) + count(version1.cbegin(), version1.cend(), ‘0‘)) == version1.length()) return 0; else return 1; } if (num1 == num2&&num1==0)return compareVersion(version1.substr(pos1+1), version2.substr(pos2+1)); if (num1 == num2&&num1 != 0)return compareVersion(version1.substr(pos1), version2.substr(pos2)); else { if (num1 > num2)return 1; if (num1 < num2)return -1; } } } return 0; } };
LeetCode 165.Compare Version Numbers
原文：http://www.cnblogs.com/csudanli/p/5744777.html