Physical Address
304 North Cardinal St.
Dorchester Center, MA 02124
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:
version1 < version2
, return -1
.version1 > version2
, return 1
.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.version1
and version2
can be stored in a 32-bit integer.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;
}
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;
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;
}
}
In our experience, we suggest you solve this Compare Version Numbers LeetCode Solution and gain some new skills from Professionals completely free and we assure you will be worth it.
If you are stuck anywhere between any coding problem, just visit Queslers to get the Compare Version Numbers LeetCode Solution
I hope this Compare Version Numbers LeetCode Solution would be useful for you to learn something new from this problem. If it helped you then don’t forget to bookmark our site for more Coding Solutions.
This Problem is intended for audiences of all experiences who are interested in learning about Data Science in a business context; there are no prerequisites.
Keep Learning!
More Coding Solutions >>