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Leetcode 165 Compare Version Numbers

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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:

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

Example 2:

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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:

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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.

Example 4:

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Input: version1 = "1.0.1", version2 = "1"
Output: 1

Example 5:

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Input: version1 = "7.5.2.4", version2 = "7.5.3"
Output: -1

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.

双指针

对于这两个字符串, 用两个双指针, 从左到右地每一次都获取一个版本号, 比较这两个版本号. 如果相等, 就比较下一个版本号.

如果某个字符串结束了, 就只比较另外一个字符串中剩余版本号和0的关系即可

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class Solution {
    public int compareVersion(String version1, String version2) {
        int begin1 = 0;
        int end1 = 0;
        int begin2 = 0;
        int end2 = 0;
        int len1 = version1.length();
        int len2 = version2.length();
        while(end1 < len1 || end2 < len2)
        {
            while(end1 < len1 && version1.charAt(end1) == '.')
            {
                ++begin1;
                ++end1;
            }
            while(end2 < len2 && version2.charAt(end2) == '.')
            {
                ++begin2;
                ++end2;
            }
            if(end1 >= len1)
            {
                while(end2 < len2 && version2.charAt(end2) != '.')
                {
                    ++end2;
                }
                int v2 = Integer.parseInt(version2.substring(begin2, end2));
                begin2 = end2 = end2 + 1;
                if(v2 > 0)
                    return -1;
            }
            else if(end2 >= len2)
            {
                while(end1 < len1 && version1.charAt(end1) != '.')
                {
                    ++end1;
                }
                int v1 = Integer.parseInt(version1.substring(begin1, end1));
                begin1 = end1 = end1 + 1;
                if(v1 > 0)
                    return 1;
            }
            else
            {
                while(end1 < len1 && version1.charAt(end1) != '.')
                {
                    ++end1;
                }
                int v1 = Integer.parseInt(version1.substring(begin1, end1));
                begin1 = end1;
                
                while(end2 < len2 && version2.charAt(end2) != '.')
                {
                    ++end2;
                }
                int v2 = Integer.parseInt(version2.substring(begin2, end2));
                begin2 = end2;
                if(v1 < v2)
                    return -1;
                else if(v1 > v2)
                    return 1;
            }
        }
        return 0;
    }
}