Leetcode 165. Compare version number

Problem

Leet code problem 165
Difficulty: Medium

Solution

1. Split each version into arrays of revisions using string.split(\\.)

    version1: 1.0.01 -> ["1", "0", "01"]
    version2: 1.0    -> ["1", "0"]
   

2. Make sure the 2 arrays has the same size, the one has smaller size has to fill remaining with 0s logically

    version1: 1.0.01 -> ["1", "0", "01"] -> ["1", "0", "01"]
    version2: 1.0    -> ["1", "0"]       -> ["1", "0", "0"]
    Logically means v2.length = 2                       ^
    then consider index from 3 onward as 0
   

3. For each item in both extracted arrays, compare and return accordingly

    version1 > version2 ->  1
    version1 = version2 ->  0
    version1 < version2 -> -1
   

Implementation

Code from github

public class L165_CompareVersionNumbers {
  public int compareVersion(String version1, String version2) {
    String[] v1 = version1.split("\\.");
    String[] v2 = version2.split("\\.");

    int length = Math.max(v1.length, v2.length);
    for (int i = 0; i < length; i++) {
      int v1Revision = i >= v1.length ? 0 : Integer.valueOf(v1[i]);
      int v2Revision = i >= v2.length ? 0 : Integer.valueOf(v2[i]);

      if (v1Revision > v2Revision) {
        return 1;
      } else if (v1Revision < v2Revision) {
        return -1;
      }
    }
    return 0;

  }
}