Leetcode 165 Compare Version Numbers

triplebee

发布于 2018-01-12 14:47:59

5040

发布于 2018-01-12 14:47:59

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

版本号比较问题。

模拟，练练手，注意边界情况，点可能有多个

class Solution {
public:
    int compareVersion(string version1, string version2) {
        if(version1.empty() && version2.empty()) return 0;
        if(version1.empty()) version1+="0";
        if(version2.empty()) version2+="0";
        int index1 = version1.find(".");
        int index2 = version2.find(".");
        int v11 = index1 == -1 ? stoi(version1) : stoi(version1.substr(0, index1));
        int v21 = index2 == -1 ? stoi(version2) : stoi(version2.substr(0, index2));
        if(v11 > v21) return 1;
        else if(v11 < v21) return -1;
        else
        {
            string s1, s2;
            if(index1 != -1) s1 = version1.substr(index1+1); 
            if(index2 != -1) s2 = version2.substr(index2+1);
            return compareVersion(s1, s2);
        }

    }
};

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原始发表：2017-01-09 ，如有侵权请联系 cloudcommunity@tencent.com 删除

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