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: String

Split the version string to a list of numbers, and compare two lists.

Time complexity: O(l1 + l2)
Space complexity: O(l1 + l2)

// Author: Huahua
class Solution {
public:
  int compareVersion(string version1, string version2) {
    const auto& v1 = parseVersion(version1);
    const auto& v2 = parseVersion(version2);
    int i1 = 0, i2 = 0;
    while (i1 < v1.size() || i2 < v2.size()) {    
      int n1 = i1 < v1.size() ? v1[i1++] : 0;
      int n2 = i2 < v2.size() ? v2[i2++] : 0;
      if (n1 < n2) return -1;
      else if (n1 > n2) return 1;
    }
    return 0;
  }
private:
  vector<int> parseVersion(const string& version) {
    vector<int> v;
    int s = 0;
    for (char c : version) {
      if (c == '.') {
        v.push_back(s);
        s = 0;
      } else {
        s = s * 10 + (c - '0');
      }
    }
    v.push_back(s);
    return v;
  }
};

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