LeetCode 165. Compare Version Numbers

题目描述:

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

使用.把字符串分割为数组, 然后再依次比较大小.

class Solution {
public:
    int compareVersion(string version1, string version2) {
        vector<int> v1 = versionVector(version1), v2 = versionVector(version2);
        for(int i = 0; i < v1.size() && i < v2.size(); i++){
            if(v1[i] > v2[i]) return 1;
            else if(v1[i] < v2[i]) return -1;
        }
        if(v1.size() > v2.size()){
            for(int i = v2.size(); i < v1.size(); i++){
                if(v1[i] != 0) return 1;
            }
        }
        else if(v1.size() < v2.size()){
            for(int i = v1.size(); i < v2.size(); i++){
                if(v2[i] != 0) return -1;
            }
        }
        return 0;
    }
    
    vector<int> versionVector(string &s){
        vector<int> version;
        int pos = 0, next = 0;
        while(next = s.find('.', pos)){
            version.push_back(stoi(s.substr(pos, next - pos)));
            if(next == string::npos) break;
            pos = next + 1;
        }
        return version;
    }
};