leetcode:Compare Version Numbers

题目描述：

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.

这个首先想到了android种版本升级的那个，可惜andorid用的是versioncode 而不是versionname是很简单的整数判断，不过也是很实际的例子。

考虑的情况首先是以分隔符“.”为例的 1中 1.1  2.1这种从前往后开始遍历时 分隔符之前就能判断出1.1是小于2.1的 return -1；

对于 1.1 和 1.2 当分隔符之前相等 此时重新判断分隔符之后内容  1.1<1.2 return -1 ；

对于更多影响的元素 譬如 0.2.1 1.1或者 1.3.1 1.3  ，1.2.0.1  1.1.2.3，等即保持从前往后遍历，遇到分隔符“.”即开始分割比较，是可以比较出结果的。

accepted code with c++:

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;   //这里用到的一个技巧即将分隔符之前已经比较出来的结果重置，无需再对temp1，temp2相加，影响后面的运算               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;    }};