165. Compare Version Numbers

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 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

Solution 1

/**
 * @param {string} version1
 * @param {string} version2
 * @return {number}
 */
var compareVersion = function(version1, version2) {
  var v1 = version1.split('.');
  var v2 = version2.split('.');
  var len = Math.max(v1.length, v2.length);
  var t1 = 0;
  var t2 = 0;
  for (var i = 0; i < len; i++) {
    t1 = parseInt(v1[i] || 0);
    t2 = parseInt(v2[i] || 0);
    if (t1 > t2) return 1;
    if (t1 < t2) return -1;
  }
  return 0;
};

Solution 2

class Solution {
    public int compareVersion(String version1, String version2) {
        // acquire first number
        int numA = 0;
        int i;
        for (i = 0; i < version1.length(); i++) {
            char c = version1.charAt(i);
            if (c == '.') {
                break;
            } else {
                numA = numA * 10 + (c - 48);
            }
        }
        // acquire second number
        int numB = 0;
        int j;
        for (j = 0; j < version2.length(); j++) {
            char c = version2.charAt(j);
            if (c == '.') {
                break;
            } else {
                numB = numB * 10 + (c - 48);
            }
        }
        // compare
        if (numA > numB) {
            return 1;
        } else if (numA < numB) {
            return -1;
        } else {
            // equal
            String v1 = "";
            String v2 = "";
            if (i != version1.length()) {
                v1 = version1.substring(i + 1);
            }
            if (j != version2.length()) {
                v2 = version2.substring(j + 1);
            }
            // if both versions end here, they are equal
            if (v1.equals("") && v2.equals("")) {
                return 0;
            } else {
                return compareVersion(v1, v2);
            }
        }
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(1).