ProblemCompare 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.You may assume the default revision number for each level of a version number to be 0. For example, version number 3.4 has a revision number of 3 and 4 for its first and second level revision number. Its third and fourth level revision number are both 0.Example 1:Input: version1 = “0.1”, version2 = “1.1"Output: -1Example 2:Input: version1 = “1.0.1”, version2 = “1"Output: 1Example 3:Input: version1 = “7.5.2.4”, version2 = “7.5.3"Output: -1Example 4:Input: version1 = “1.01”, version2 = “1.001"Output: 0Explanation: Ignoring leading zeroes, both “01” and “001” represent the same number “1”Example 5:Input: version1 = “1.0”, version2 = “1.0.0"Output: 0Explanation: The first version number does not have a third level revision number, which means its third level revision number is default to “0"Note:Version strings are composed of numeric strings separated by dots . and this numeric strings may have leading zeroes.Version strings do not start or end with dots, and they will not be two consecutive dots.Solutionclass 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++) { int n1 = i < v1.length ? Integer.parseInt(v1[i]) : 0; int n2 = i < v2.length ? Integer.parseInt(v2[i]) : 0; int ans = Integer.compare(n1, n2); if (ans != 0) return ans; } return 0; }}