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  • 165. Compare Version Numbers (String)

    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

    注意:版本格式可能会有多个子版本

    class Solution {
    public:
        int compareVersion(string version1, string version2) {
            vector<int> arr_v1,arr_v2;
            int pos1_pre = 0, pos2_pre = 0, pos1=version1.find_first_of('.'),pos2=version2.find_first_of('.');
            while(pos1!=-1){
                arr_v1.push_back(atoi(version1.substr(pos1_pre,pos1).c_str()));
                pos1_pre = pos1+1;
                pos1 = version1.find_first_of('.',pos1_pre);
            }
            arr_v1.push_back(atoi(version1.substr(pos1_pre).c_str()));
            
            while(pos2!=-1){
                arr_v2.push_back(atoi(version2.substr(pos2_pre,pos2).c_str()));
                pos2_pre = pos2+1;
                pos2 = version2.find_first_of('.',pos2_pre);
            }
            arr_v2.push_back(atoi(version2.substr(pos2_pre).c_str()));
            
            int size1 = arr_v1.size();
            int size2 = arr_v2.size();
            int size = min(size1,size2);
            for(int i = 0; i < size; i++){
                if(arr_v1[i] > arr_v2[i]) return 1;
                else if(arr_v1[i] < arr_v2[i]) return -1;
            }
            if(size == size1){
                for(int i = size; i < size2; i++){
                    if(arr_v2[i]!=0) return -1;
                }
                return 0;
            }
            else{
                for(int i = size; i < size1; i++){
                    if(arr_v1[i]!=0) return 1;
                }
                return 0;
            }
        }
    };
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  • 原文地址:https://www.cnblogs.com/qionglouyuyu/p/6602298.html
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