[leetcode #165] Compare Version Numbers

Problem

Given two version numbers, version1 and version2, compare them.

Version numbers consist of one or more revisions joined by a dot '.'. Each revision consists of digits and may contain leading zeros. Every revision contains at least one character. Revisions are 0-indexed from left to right, with the leftmost revision being revision 0, the next revision being revision 1, and so on. For example 2.5.33 and 0.1 are valid version numbers.

To compare version numbers, compare their revisions in left-to-right order. Revisions are compared using their integer value ignoring any leading zeros. This means that revisions 1 and 001 are considered equal. If a version number does not specify a revision at an index, then treat the revision as 0. For example, version 1.0 is less than version 1.1 because their revision 0s are the same, but their revision 1s are 0 and 1 respectively, and 0 < 1.

Return the following:

・ If version1 < version2, return -1.
・ If version1 > version2, return 1.
・ Otherwise, return 0.

Example 1:

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

Example 2:

Input: version1 = "1.0", version2 = "1.0.0"
Output: 0
Explanation: version1 does not specify revision 2, which means it is treated as "0".

Example 3:

Input: version1 = "0.1", version2 = "1.1"
Output: -1
Explanation: version1's revision 0 is "0", while version2's revision 0 is "1". 0 < 1, so version1 < version2.

Constraints:

・ 1 <= version1.length, version2.length <= 500
・ version1 and version2 only contain digits and '.'.
・ version1 and version2 are valid version numbers.
・ All the given revisions in version1 and version2 can be stored in a 32-bit integer.

Idea

주어진 버전 중 최신 버전이 무엇인지 구하는 문제다.

버전의 각 revision은 '.'으로 구분되어 있으며 각 revision 중 상위 revision이 더 크면 버전이 최신 버전이다.

'.'의 index를 찾아 각 revision을 integer로 바꾼 뒤 비교해서 풀어도 되고, '.'를 기준으로 revision의 string을 array로 만들어 array의 element를 비교하는 방법이 있다.

substring으로 푸는 방법이 좀 더 빠르긴 하지만 마지막 revision을 처리하는 게 코드가 더 들어가서 길어지는 면이 있다.

Time Complexity: O(n)
Space Complexity: O(1)

Solution

substring으로 푼 답. (코드가 더럽다)

class Solution {
    public int compareVersion(String version1, String version2) {
        int dotIndex1 = 0;
        int dotIndex2 = 0;
        
        while (true) {
            int index1 = version1.indexOf('.', dotIndex1);
            int index2 = version2.indexOf('.', dotIndex2);

            int val1 = 0;
            if (index1 != -1) {
                val1 = Integer.parseInt(version1.substring(dotIndex1, index1));
                dotIndex1 = index1+1;
            } else if (dotIndex1 == version1.length()) {
                val1 = 0;
            } else {
                val1 = Integer.parseInt(version1.substring(dotIndex1));
                dotIndex1 = version1.length();
            }

            int val2 = 0;
            if (index2 != -1) {
                val2 = Integer.parseInt(version2.substring(dotIndex2, index2));
                dotIndex2 = index2+1;
            } else if (dotIndex2 == version2.length()) {
                val2 = 0;
            } else {
                val2 = Integer.parseInt(version2.substring(dotIndex2));
                dotIndex2 = version2.length();
            }

            if (val1 > val2) {
                return 1;
            }

            if (val1 < val2) {
                return -1;
            }

            if (index1 == -1 && index2 == -1) {
                break;
            }
        }

        return 0;
    }
}

split으로 해서 푼 답

class Solution {
    public int compareVersion(String version1, String version2) {
        String[] str1 = version1.split("\\.");
        String[] str2 = version2.split("\\.");
        int max = Math.max(str1.length,str2.length);
        for(int i=0;i<max;i++){
            int num1 = i >= str1.length ? 0 : Integer.parseInt(str1[i]);
            int num2 = i >= str2.length ? 0 : Integer.parseInt(str2[i]);
            if(num1 < num2) return -1;
            if(num1 > num2) return 1;
        }

        return 0;
    }
}

Reference

https://leetcode.com/problems/compare-version-numbers/