[LeetCode] 165. Compare Version Numbers - 문제풀이

Description

주어진 두개의 문자열의 버전을 비교하여 반환하는 문제입니다.

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.

Solution 1. String

public int compareVersion(String version1, String version2) {

    String[] v1 = version1.split("\\\\.");
    String[] v2 = version2.split("\\\\.");
    int i = 0;
    while(i < Math.max(v1.length,v2.length) ){
        int n1 = v1.length > i? Integer.valueOf(v1[i]) : 0;
        int n2 = v2.length > i? Integer.valueOf(v2[i]) : 0;
        if(n1 != n2) {
            return n1 > n2? 1 : -1;
        }
        i++;
    }
    return 0;
}

점(.)으로 스플릿하여 앞에서 부터 비교하고 값이 같을 경우 다음 하위 버전을 비교 할 수 있도록 반복하며 비교해줍니다.

Reference

Compare Version Numbers - LeetCode