题目

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

分析

方法一：

采用字符串比较；此题中的版本比较，需要将.号前的部分与第二部分分开比较；

方法二：

将版本号的字符串转化为整数；

AC代码

class Solution {

public:

    /*方法一：字符串比较；如果version1 > version2 则返回1 相等返回0 反之返回-1*/

    int compareVersion1(string version1, string version2) {

        if (version1.empty() && version2.empty())

            return 0;

        else if (version2.empty())

            return 1;

        else if (version1.empty())

            return -1;

        else

        {

            //分别得到版本号的第一部分和第二部分

            int pos = 0;

            for (int i = 0; version1[i] != '\0' && version1[i] != '.'; ++i)

            {

                ++pos;

            }//for

            string fir_version1;

            string sec_version1;

            if (pos == version1.length())

            {

                fir_version1 = version1;

                sec_version1 = "";

            }

            else{

                fir_version1 = version1.substr(0, pos);

                sec_version1 = version1.substr(pos + 1, version1.length() - pos - 1);

            }

            //要略过第一部分版本号的开头的0

            pos = 0;

            while (fir_version1[pos] != '\0' && fir_version1[pos] == '0')

            {

                ++pos;

            }//while

            if (fir_version1[pos] == '\0')

                fir_version1 = "0";

            else

                fir_version1 = fir_version1.substr(pos, fir_version1.length() - pos);

            pos = 0;

            for (int i = 0; version2[i] != '\0' && version2[i] != '.'; ++i)

            {

                ++pos;

            }//for

            string fir_version2;

            string sec_version2;

            if (pos == version2.length())

            {

                fir_version2 = version2;

                sec_version2 = "";

            }

            else{

                fir_version2 = version2.substr(0, pos);

                sec_version2 = version2.substr(pos + 1, version2.length() - pos - 1);

            }//else

            pos = 0;

            while (fir_version2[pos] != '\0' && fir_version2[pos] == '0')

            {

                ++pos;

            }//while

            if (fir_version2[pos] == '\0')

                fir_version2 = "0";

            else

                fir_version2 = fir_version2.substr(pos, fir_version2.length() - pos);

            if (strcmp(fir_version1.c_str(), fir_version2.c_str()) != 0)

                return strcmp(fir_version1.c_str(), fir_version2.c_str());

            else

                return strcmp(sec_version1.c_str(), sec_version2.c_str());

        }

    }

    //方法二：化为整数比较

    int compareVersion(string version1, string version2) {

        int val1, val2;

        int idx1 = 0, idx2 = 0;

        while (idx1 < version1.length() || idx2 < version2.length()) {

            val1 = 0;

            while (idx1 < version1.length()) {

                if (version1[idx1] == '.') {

                    ++idx1;

                    break;

                }

                val1 = val1 * 10 + (version1[idx1] - '0');

                ++idx1;

            }

            val2 = 0;

            while (idx2 < version2.length()) {

                if (version2[idx2] == '.') {

                    ++idx2;

                    break;

                }

                val2 = val2 * 10 + (version2[idx2] - '0');

                ++idx2;

            }

            if (val1 > val2) return 1;

            if (val1 < val2) return -1;

        }

        return 0;

    }

};

GitHub测试程序源码