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:

  1. Version strings are composed of numeric strings separated by dots . and this numeric strings may have leading zeroes.
  2. Version strings do not start or end with dots, and they will not be two consecutive dots.

Solution

Time complexity : O(n)
Space complexity : O(n)

class Solution {
public:
    int compareVersion(string version1, string version2) {
        vector<int> v1 = parseVer(version1);
        vector<int> v2 = parseVer(version2);
        
        int idx = 0;
        int size1 = v1.size();
        int size2 = v2.size();
        while (idx < max(size1, size2)) {
            int n1 = idx < size1 ? v1[idx] : 0;
            int n2 = idx < size2 ? v2[idx] : 0;
            ++idx;
            if (n1 < n2) return -1;
            else if (n1 > n2) return 1;
        }
        return 0;
    }
private:
    vector<int> parseVer(string& str) {
        vector<int> res;
        int sum = 0;
        for(char c: str) {
            if (c == '.') {
                res.push_back(sum);
                sum = 0;
            } else {
                sum = 10*sum + (c-'0');
            }
        }
        res.push_back(sum);
        return res;
    }
};

先個別用vector存放version1、version2，以「.」隔開的版本號。
再從頭一個一個比較，超過的用0代替，照題目所述輸出。