Compare Version Numbers

Total Accepted: 39335 Total Submissions: 242429 Difficulty: Easy

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
class Solution {
public:
    int strToInt(string& s,int start,int end)
    {
        int res = 0;
        while(start<end){
            res = res*10+(s[start++]-'0');
        }
        return res;
    }
    int compareVersion(string version1, string version2) {
        int v1size = version1.size();
        int v2size = version2.size();
        int i=0,j=0,starti=0,startj=0;
        int v1=0,v2=0;
        while(i<v1size && j<v2size){
            while(i<v1size && version1[i]!='.') i++;
            while(j<v2size && version2[j]!='.') j++;
            v1 = i==0 ? -1 :strToInt(version1,starti,i);
            v2 = j==0 ? -1 :strToInt(version2,startj,j);
            if(v1<v2){
                return -1;
            }else if(v1>v2){
                return 1;
            }else{
                starti = ++i;
                startj = ++j;
            }
        }
        while(i<v1size && (version1[i]=='0' || version1[i]=='.')) i++;
        while(j<v2size && (version2[j]=='0' || version2[j]=='.')) j++;
        if(i<v1size){
            return 1;
        }
        if(j<v2size){
            return -1;
        }
        return 0;
    }
};
原文地址:https://www.cnblogs.com/zengzy/p/5036507.html