LeetCode-Compare Version Numbers

LeetCode-Compare Version Numbers

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

Credits:

Special thanks to @ts for adding this problem and creating all test cases.

結題報告：題目給的例子，說的不太明白，版本號還有1.2.4 01.2.4.3這種類型的。所以只要搜索'.'。再進行判斷就行

class Solution {
public:
    int compareVersion(string version1, string version2) {
      int temp1 = 0;
      int temp2 = 0;
      string::iterator t1 = version1.begin();
      string::iterator t2 = version2.begin();
      while(t1 != version1.end() || t2 != version2.end())
      {
        while(*t1 != '.' && t1 != version1.end())
        {
          temp1 = temp1 *10 + *t1++ - '0';

        }

        while(*t2 != '.' && t2 != version2.end())
        {
          temp2 = temp2 *10 +  *t2++ - '0';

        }
        if(temp1 > temp2)    return 1;
        else if(temp1 < temp2) return -1;
        else {
          if(t1 == version1.end() && t2 == version2.end())
            return 0;
          else if(t1 == version1.end() && t2 != version2.end()) {
               temp1 = 0;
               temp2 = 0;
               t2++;
               }
          else if(t1 != version1.end() && t2 == version2.end()) {
               temp1 = 0;
               temp2 = 0;
               t1++;
              }
          else {
                temp1 = 0;
                temp2 = 0;
                t1++;
                t2++;
               }
            }
      }
      return 0;
    }
};