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# Compare Version Numbers LeetCode Solution

## Problem – Compare Version Numbers LeetCode Solution

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.

## Compare Version Numbers LeetCode Solution in Java

``````public int compareVersion(String version1, String version2) {
String[] levels1 = version1.split("\\.");
String[] levels2 = version2.split("\\.");

int length = Math.max(levels1.length, levels2.length);
for (int i=0; i<length; i++) {
Integer v1 = i < levels1.length ? Integer.parseInt(levels1[i]) : 0;
Integer v2 = i < levels2.length ? Integer.parseInt(levels2[i]) : 0;
int compare = v1.compareTo(v2);
if (compare != 0) {
return compare;
}
}

return 0;
}
``````

## Compare Version Numbers LeetCode Solution in Python

``````def compareVersion(self, version1, version2):
versions1 = [int(v) for v in version1.split(".")]
versions2 = [int(v) for v in version2.split(".")]
for i in range(max(len(versions1),len(versions2))):
v1 = versions1[i] if i < len(versions1) else 0
v2 = versions2[i] if i < len(versions2) else 0
if v1 > v2:
return 1
elif v1 < v2:
return -1;
return 0;
``````

## Compare Version Numbers LeetCode Solution in C++

``````int compareVersion(string version1, string version2) {
for(auto& w : version1) if (w == '.') w=' ';
for(auto& w : version2) if (w == '.') w=' ';
istringstream s1(version1), s2(version2);
while(1) {
int n1,n2;
if (not(s1 >> n1) ) n1 = 0;
if (not(s2 >> n2) ) n2 = 0;
if (not s1 and not s2) return 0;
if (n1<n2) return -1;
if (n1>n2) return 1;
}
}
``````
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