[leetcode 165] Compare Version Numbers

最新推荐文章于 2022-02-25 21:50:14 发布

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本文介绍了一种用于比较两个版本号大小的算法实现。通过解析版本号字符串，逐段比较每部分数字来确定版本间的大小关系。该算法适用于软件版本管理和自动化测试等场景。

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) {
        const int n = version1.size();
        const int m = version2.size();
        int v1 = 0;
        int v2 = 0;
        int i = 0;
        int j = 0;
        while (i < n && j < m) {
            for (; i < n; i++) {
                if (version1[i] == '.') {
                    i = i + 1;
                    break;
                }
                v1 = v1*10 + (version1[i] - '0');
            }
            for (; j < m; j++) {
                if (version2[j] == '.') {
                    j = j + 1;
                    break;
                }
                v2 = v2*10 + (version2[j] - '0');
            }
            if (v1 < v2) {
                return -1;
            } else if (v1 > v2) {
                return 1;
            }
            v1 = v2 = 0;
        }
        if (i == n && j == m) {
            return 0;
        }
        if (i < n) {
            while (i < n) {
                if (version1[i] != '0' && version1[i] != '.') {
                    return 1;
                }
                i++;
            }
            return 0;
        } else {
            while (j < m) {
                if (version2[j] != '0' && version2[j] != '.') {
                    return -1;
                }
                j++;
            }
            return 0;
        }
    }
};