Compare Version Numbers--LeetCode

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

思路：每次比较小数点之前的数，看能不能分出伯仲。就是比较“版本号”大小，给定的版本号version1和version2是字符串类型的，当version1>version2的时候，返回1，反之返回-1。

这道题属于细节处理题，除了字符串处理繁琐一点之外没有什么。

解题思路：先分别将version1、version2字符串按'.'分割成多个子串，每个子串转化成整型存入容器。最后再比较两个容器中对应位置的数的大小。当然需要考虑它们长度不同的情况。

注意点：

（1）version的形式：10.23.1、01.23、1.2.3.4....诸如

int compareVersion(string& s1, string& s2) {
    int i=0,j=0;
    int temp1,temp2;
    while(i<s1.length() && j<s2.length())
    {
       temp1=0;
       temp2=0;
       for(;i<s1.length();i++)
         if(s1[i]!='.')
         temp1  = temp1*10 + (s1[i]-'0');
       for(;j<s2.length();j++)
         if(s2[j]!='.')
         temp2  = temp2*10 + (s2[j]-'0');
       if(temp1> temp2)
        return 1;
       else if(temp1<temp2)
        return -1;
       else
       {
           i++;
           j++;
       }                  
    }    
    if(i>=s1.length() && j< s2.length())
      return -1;
    if(i<s1.length() && j>=s2.length())
      return 1;
    if(i>=s1.length() && j>=s2.length())
      return 0;
}