Algorithm
Problem Name: 165. Compare Version Numbers
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, return1. - 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 <= 500version1andversion2only contain digits and'.'.version1andversion2are valid version numbers.- All the given revisions in
version1andversion2can be stored in a 32-bit integer.
Code Examples
#1 Code Example with C Programming
Code -
C Programming
int compareVersion(char* version1, char* version2) {
int a, b;
char *v1, *v2;
while (version1 || version2) {
if (version1) {
v1 = version1;
version1 = strchr(version1, '.');
if (version1) {
*version1 = 0;
version1 ++;
}
a = atoi(v1);
} else {
a = 0;
}
if (version2) {
v2 = version2;
version2 = strchr(version2, '.');
if (version2) {
*version2 = 0;
version2 ++;
}
b = atoi(v2);
} else {
b = 0;
}
if (a > b) return 1;
if (a < b) return -1;
}
return 0;
}
Input
#2 Code Example with Java Programming
Code -
Java Programming
class Solution {
public int compareVersion(String version1, String version2) {
String[] splitOne = version1.split("\\.");
String[] splitTwo = version2.split("\\.");
int i = 0;
for (i = 0; i < Math.min(splitOne.length, splitTwo.length); i++) {
int diff = Integer.parseInt(splitOne[i]) - Integer.parseInt(splitTwo[i]);
if (diff == 0) {
continue;
}
return diff < 0 ? -1 : 1;
}
while (i < splitOne.length) {
if (Integer.parseInt(splitOne[i++]) > 0) {
return 1;
}
}
while (i < splitTwo.length) {
if (Integer.parseInt(splitTwo[i++]) > 0) {
return -1;
}
}
return 0;
}
}
Input
#3 Code Example with Javascript Programming
Code -
Javascript Programming
start coding...Input
#4 Code Example with Python Programming
Code -
Python Programming
class Solution:
def compareVersion(self, version1, version2):
def getNum(s):
if not s: return (None, None)
for i in range(len(s)):
if s[i] == ".": return (s[i + 1:], int(s[:i]))
return (None, int(s))
while True:
version1, n1 = getNum(version1)
version2, n2 = getNum(version2)
if version1 == version2 == n1 == n2 == None: return 0
if n1 != None and n1 > 0 and (n2 == None or n1 > n2): return 1
if n2 != None and n2 > 0 and (n1 == None or n2 > n1): return -1
Input
#5 Code Example with C# Programming
Code -
C# Programming
using System;
namespace LeetCode
{
public class _0165_CompareVersionNumbers
{
public int CompareVersion(string version1, string version2)
{
var nums1 = version1.Split(new char[] { '.' });
var nums2 = version2.Split(new char[] { '.' });
var length1 = nums1.Length;
var length2 = nums2.Length;
for (int i = 0; i < Math.Max(length1, length2); i++)
{
var current1 = i < length1 ? int.Parse(nums1[i]) : 0;
var current2 = i < length2 ? int.Parse(nums2[i]) : 0;
if (current1 != current2)
return current1 > current2 ? 1 : -1;
}
return 0;
}
}
}
Input
Output