Leetcode 165. Compare Version Numbers

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

思路:
比较版本号。
注意版本号可能的坑:
1.0和1.0.0 1.0 == 1.0.0
0.01和0.0.1 0.01 > 0.0.1

public int compareVersion(String version1, String version2) {
    //String[] v1s = version1.split(".");//bug
    //String[] v2s = version2.split(".");
    String[] v1s = version1.split("\\.");
    String[] v2s = version2.split("\\.");

    int maxLen = Math.max(v1s.length, v2s.length);
    for (int i = 0; i < maxLen; i++) {
        int v1 = i < v1s.length ? Integer.valueOf(v1s[i]) : 0;
        int v2 = i < v2s.length ? Integer.valueOf(v2s[i]) : 0;
        if (v1 > v2) {
            return 1;
        } else if (v1 < v2) {
            return -1;
        }
    }

    return 0;
}
    原文作者:ShutLove
    原文地址: https://www.jianshu.com/p/bb90b4b58309
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