学步园

题目

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

int compareVersion(string version1, string version2){

}

题目大意

比较两个版本数version1和version2。如果version1 > version2 返回 1, version < verison2 返回 2， 否则返回0；

假设版本字符串非空，只包含'.'和数字。

难度系数: 容易

实现

int compareVersion(string version1, string version2) {
    int ver1, ver2;
    if (version1.empty() && version2.empty())
        return 0;
    size_t pos1 = version1.find(".");
    if (version1.empty()) {
        ver1 = 0;
    } else {
        if (pos1 == string::npos) {
            pos1 = version1.length();
        }
        ver1 = atoi(version1.substr(0, pos1).c_str());
    }
    size_t pos2 = version2.find(".");
    if (version2.empty()) {
        ver2 = 0;
    } else {
        if (pos2 == string::npos) {
            pos2 = version2.length();
        }
        ver2 = atoi(version2.substr(0, pos2).c_str());
    }
    if (ver1 > ver2)
        return 1;
    else if (ver1 < ver2)
        return -1;
    else
        return compareVersion(pos1 == version1.length()? "" : version1.substr(pos1+1), pos2 == version2.length() ? "" : version2.substr(pos2+1));
}

比较版本号，这个程序是比较实用的，实现并不难，Java的字符串有个split方法，实现这个应该更容易些。