165-compare-version-numbers

Question

https://leetcode.com/problems/compare-version-numbers/description/

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:

0.1 < 1.1 < 1.2 < 13.37

Thought Process

1. Split and Compare
   1. Start from the first field until we reach the end or the numbers are different
   2. Time complexity O(n), where n is number of fields
   3. Space complexity O(n)

Solution

class Solution {
    public int compareVersion(String version1, String version2) {
        String[] ver1 = version1.split("\\.");
        String[] ver2 = version2.split("\\.");
        int n = Math.max(ver1.length, ver2.length);
        for (int i = 0; i < n; i++){
            int v1 = i >= ver1.length ? 0 : Integer.parseInt(ver1[i]);
            int v2 = i >= ver2.length ? 0 : Integer.parseInt(ver2[i]);
            int cmp = Integer.compare(v1, v2);
            if (cmp != 0) return cmp;
        }
        return 0;
    }
}

Additional