[LeetCode] Compare Version Numbers

Alex 發布于2019-08-15 15:51 / 2160人閱讀

Problem

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.

Example

Example 1:

Input: version1 = "0.1", version2 = "1.1"
Output: -1

Example 2:

Input: version1 = "1.0.1", version2 = "1"
Output: 1

Example 3:

Input: version1 = "7.5.2.4", version2 = "7.5.3"
Output: -1

Note

https://stackoverflow.com/que...
Basically if you want to split a dot ".", the regex "." means "any character", so you need to escape the dot with two backsplashes.

Solution

class Solution {
    public int compareVersion(String version1, String version2) {
        String[] v1 = version1.split(".");
        String[] v2 = version2.split(".");
        int len = Math.max(v1.length, v2.length);
        for (int i = 0; i < len; i++) {
            Integer i1 = i < v1.length ? Integer.parseInt(v1[i]) : 0;
            Integer i2 = i < v2.length ? Integer.parseInt(v2[i]) : 0;
            int compare = i1.compareTo(i2);
            if (compare != 0) return compare;
        }
        return 0;
    }
}

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