leetcode 165. Compare Version Numbers

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


题意非常明白，就是按照version比较规则来比较。
自己没怎么想，看了别人的解法。

碰到‘.’之前一直累加数，碰到'.'进行比较，直至字符串结束。
字符串转数字s[i]-'0'。
就没有什么难点了。

class Solution {
public:

    int compareVersion(string version1, string version2) {
        int val1=0,val2=0;
        int len1=version1.length();
        int len2=version2.length();
        int i=0,j=0;
        
        while(i<len1||j<len2){
            val1=0;val2=0;
            while(i<len1){
            if(version1[i]=='.')  {i++;break;}
            else  val1=val1*10+version1[i++]-'0';}
            while(j<len2){
            if(version2[j]=='.')  {j++;break;}
            else val2=val2*10+version2[j++]-'0';}
            
            if(val1>val2) return 1;
            if(val1<val2) return -1;
        }        
        return 0;
        
    }
};