165. Compare Version Numbers

• Difficulty: Medium

• Topics: String

• Similar Questions:

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.

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: -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

Example 4:

Input: version1 = "1.01", version2 = "1.001"
Output: 0
Explanation: Ignoring leading zeroes, both “01” and “001" represent the same number “1”

Example 5:

Input: version1 = "1.0", version2 = "1.0.0"
Output: 0
Explanation: The first version number does not have a third level revision number, which means its third level revision number is default to "0"

Note:

Solutions:

class Solution {
public:
    int compareVersion(string version1, string version2) {
        int cur1 = 0;
        int cur2 = 0;
        int len1 = version1.length();
        int len2 = version2.length();

        while (cur1 < len1 || cur2 < len2) {
            int value1 = 0;
            int value2 = 0;

            while (cur1 < len1 && version1[cur1] == '0') {
                ++cur1;
            }
            while (cur1 < len1 && version1[cur1] != '.') {
                value1 = 10 * value1 + version1[cur1] - '0';
                ++cur1;
            }

            if (cur1 != len1) {
                ++cur1;
            }

            while (cur2 < len2 && version2[cur2] == '0') {
                ++cur2;
            }
            while (cur2 < len2 && version2[cur2] != '.') {
                value2 = 10 * value2 + version2[cur2] - '0';
                ++cur2;
            }

            if (cur2 != len2) {
                ++cur2;
            }

            if (value1 < value2) {
                return -1;
            } else if (value1 > value2) {
                return 1;
            }
        }

        return 0;

    }
};