[LeetCode] Compare Version Numbers-优快云博客

本文介绍了一种用于比较两个软件版本号大小的算法实现。通过解析版本号字符串，并逐位比较每个数字部分，来确定版本之间的先后顺序。适用于版本号由数字和点组成的情况。

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Compare two version numbers version1 and version1.
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 compareVersion(string version1, string version2) {
        int v1 = 0,v2 = 0,idx1 = 0,idx2 = 0;
        while(idx1 != version1.size() || idx2 != version2.size()){
            v1 = 0;
            while(idx1 != version1.size()){
                if(version1[idx1] == '.'){
                    idx1 ++;
                    break;
                }
                v1 = 10 * v1 + (version1[idx1] - '0');
                idx1 ++;
            }
            v2 = 0;
            while(idx2 != version2.size()){
                if(version2[idx2] == '.'){
                    idx2 ++;
                    break;
                }
                v2 = 10 * v2 + (version2[idx2] - '0');
                idx2 ++;
            }
            if(v1 > v2) return 1;
            if(v1 < v2) return -1;
        }
        return 0;
    }
};