Leetcode No.165

Compare two version numbers version1 and version2.
If version1> version2 returns 1 if version1 <version2 returns -1, except that 0 is returned.

You can assume that the version string is not empty, and contain numbers and Character.

 The character does not represent a decimal point, but used to separate sequence of numbers.

For example, 2.5 is not "two and a half" nor "bad half to three," but the second edition of the fifth small version.

You can assume that the default revision version number of each level is zero. For example, a first stage (large) versions of the 3.4 and the second stage (iteration) Amendment No. 3 and 4, respectively. The third and fourth stages of its revision number are both zero.
 

Example 1:

Input: version1 = "0.1", version2 = "1.1"
Output: -1
Example 2:

Input: version1 = "1.0.1", version2 = "1"
Output: 1
Example 3:

Input: version1 = "7.5.2.4", version2 = "7.5.3"
Output: -1
Example 4:

Input: version1 = "1.01", version2 = "1.001"
Output: 0
Explanation: ignoring leading zeros, "01" and "001" refer to the same number "a."
Example 5:

Input: version1 = "1.0", version2 = "1.0.0"
Output: 0
to explain: version1 no third-level revision number, which means that its third-level revision number defaults to "0."
 

prompt:

Version string by a dot (.) Separated numeric strings. This number string may have leading zeros.
Version string does not start or end point, and wherein no two consecutive points.

Source: stay button (LeetCode)
link: https: //leetcode-cn.com/problems/compare-version-numbers
copyrighted by deduction from all networks. Commercial reprint please contact the authorized official, non-commercial reprint please indicate the source.

Reference blog: https://www.cnblogs.com/grandyang/p/4244123.html

I initially employed using find () and substr () function to solve, and then use atoi () is converted into a digital comparison. The basic solution is consistent with the above blog, but a little tedious point.

//165
int compareVersion(string version1, string version2)
{
    if(version1.empty()) return version2.empty()?0:-1;
    if(version2.empty()) return version1.empty()?0:1;
    size_t i=0,j=0;
    while(i<version1.size() || j<version2.size())
    {
        string str1,str2;
        while(i<version1.size() && version1[i]!='.')
        {
            str1+=version1[i];
            i++;
        }
        i++;
        while(j<version2.size() && version2[j]!='.')
        {
            str2+=version2[j];
            j++;
        }
        j++;
        int num1 = atoi(str1.c_str());
        int num2 = atoi(str2.c_str());
        if(num1>num2) return -1;
        if(num1<num2) return -1;
    }
    return 0;
}//165