比较两个版本号

版本号是一个字符串，用于标识软件产品的唯一状态。版本号看起来像 abcd，其中 a、b 等是数字，所以版本号是一个字符串，其中的数字用点分隔。这些数字通常代表从主要到次要的层次结构（a 是主要的，d 是次要的）。
在这个问题中，我们得到了两个版本号。我们需要对它们进行比较，得出哪个是最新的版本号（即哪个版本号更大）。

例子：

Input: 
V1 = “1.0.31”
V2 = “1.0.27”
Output:  v1 is latest
Because V2 < V1

Input: 
V1 = “1.0.10”
V2 = “1.0.27”
Output:  v2 is latest
Because V1 < V2 

方法：由于是点，无法直接比较，但版本可以比较数字部分，然后可以找到最新版本。遍历字符串并分离数字部分并进行比较。如果相等，则比较下一个数字部分，依此类推，直到它们不同，否则将它们标记为相等。
实现一个方法来比较两个版本。如果有两个以上的版本，那么下面的 versionCompare 方法可以作为 sort 方法的比较方法，它会根据指定的比较对所有版本进行排序。

C++

// C/C++ program to compare
// two version number
#include 
using namespace std;
 
// Method to compare two versions.
// Returns 1 if v2 is smaller, -1
// if v1 is smaller, 0 if equal
int versionCompare(string v1, string v2)
{
    // vnum stores each numeric
    // part of version
    int vnum1 = 0, vnum2 = 0;
 
    // loop until both string are
    // processed
    for (int i = 0, j = 0; (i < v1.length()
                            || j < v2.length());) {
        // storing numeric part of
        // version 1 in vnum1
        while (i < v1.length() && v1[i] != '.') {
            vnum1 = vnum1 * 10 + (v1[i] - '0');
            i++;
        }
 
        // storing numeric part of
        // version 2 in vnum2
        while (j < v2.length() && v2[j] != '.') {
            vnum2 = vnum2 * 10 + (v2[j] - '0');
            j++;
        }
 
        if (vnum1 > vnum2)
            return 1;
        if (vnum2 > vnum1)
            return -1;
 
        // if equal, reset variables and
        // go for next numeric part
        vnum1 = vnum2 = 0;
        i++;
        j++;
    }
    return 0;
}
 
// Driver method to check above
// comparison function
int main()
{
    string version1 = "1.0.3";
    string version2 = "1.0.7";
 
    if (versionCompare(version1, version2) < 0)
        cout << version1 << " is smaller\n";
    else if (versionCompare(version1, version2) > 0)
        cout << version2 << " is smaller\n";
    else
        cout << "Both version are equal\n";
    return 0;
}

Java

// Java program to compare two version number
import java.util.*;
 
class GFG {
 
    // Method to compare two versions.
    // Returns 1 if v2 is
    // smaller, -1 if v1 is smaller, 0 if equal
    static int versionCompare(String v1, String v2)
    {
        // vnum stores each numeric part of version
        int vnum1 = 0, vnum2 = 0;
 
        // loop until both String are processed
        for (int i = 0, j = 0; (i < v1.length()
                                || j < v2.length());) {
            // Storing numeric part of
            // version 1 in vnum1
            while (i < v1.length()
                   && v1.charAt(i) != '.') {
                vnum1 = vnum1 * 10
                        + (v1.charAt(i) - '0');
                i++;
            }
 
            // storing numeric part
            // of version 2 in vnum2
            while (j < v2.length()
                   && v2.charAt(j) != '.') {
                vnum2 = vnum2 * 10
                        + (v2.charAt(j) - '0');
                j++;
            }
 
            if (vnum1 > vnum2)
                return 1;
            if (vnum2 > vnum1)
                return -1;
 
            // if equal, reset variables and
            // go for next numeric part
            vnum1 = vnum2 = 0;
            i++;
            j++;
        }
        return 0;
    }
 
    // Driver method to check above
    // comparison function
    public static void main(String[] args)
    {
        String version1 = "1.0.3";
        String version2 = "1.0.7";
 
        if (versionCompare(version1, version2) < 0)
            System.out.println(version1 + " is smaller");
        else if (versionCompare(version1, version2) > 0)
            System.out.println(version2 + " is smaller");
        else
            System.out.println("Both version are equal");
    }
}
 
// This code is contributed by shivanisinghss2110

Python3

# Python program to compare two version number
 
# Method to compare two versions.
# Return 1 if v2 is smaller,
# -1 if v1 is smaller,,
# 0 if equal
def versionCompare(v1, v2):
     
    # This will split both the versions by '.'
    arr1 = v1.split(".")
    arr2 = v2.split(".")
    n = len(arr1)
    m = len(arr2)
     
    # converts to integer from string
    arr1 = [int(i) for i in arr1]
    arr2 = [int(i) for i in arr2]
  
    # compares which list is bigger and fills
    # smaller list with zero (for unequal delimiters)
    if n>m:
      for i in range(m, n):
         arr2.append(0)
    else if m>n:
      for i in range(n, m):
         arr1.append(0)
     
    # returns 1 if version 1 is bigger and -1 if
    # version 2 is bigger and 0 if equal
    for i in range(len(arr1)):
      if arr1[i]>arr2[i]:
         return 1
      else if arr2[i]>arr1[i]:
         return -1
    return 0
 
# Driver program to check above comparison function
version1 = "1.0.3"
version2 = "1.0.7"
 
ans = versionCompare(version1, version2)
if ans < 0:
    print (version1 + " is smaller")
else if ans > 0:
    print (version2 + " is smaller")
else:
    print ("Both versions are equal")
 
# This code is contributed by Nikhil Kumar Singh(nickzuck_007)
# and improved by Tuhin Das (tuhindas221b)

C#

// C# program to compare
// two version number
using System;
class GFG
{
 
  // Method to compare two versions.
  // Returns 1 if v2 is smaller, -1
  // if v1 is smaller, 0 if equal
  static int versionCompare(string v1, string v2)
  {
    // vnum stores each numeric
    // part of version
    int vnum1 = 0, vnum2 = 0;
 
    // loop until both string are
    // processed
    for (int i = 0, j = 0; (i < v1.Length || j < v2.Length);)
    {
       
      // storing numeric part of
      // version 1 in vnum1
      while (i < v1.Length && v1[i] != '.')
      {
        vnum1 = vnum1 * 10 + (v1[i] - '0');
        i++;
      }
 
      // storing numeric part of
      // version 2 in vnum2
      while (j < v2.Length && v2[j] != '.')
      {
        vnum2 = vnum2 * 10 + (v2[j] - '0');
        j++;
      }
 
      if (vnum1 > vnum2)
        return 1;
      if (vnum2 > vnum1)
        return -1;
 
      // if equal, reset variables and
      // go for next numeric part
      vnum1 = vnum2 = 0;
      i++;
      j++;
    }
    return 0;
  }
 
  // Driver code
  static void Main()
  {
    string version1 = "1.0.3";
    string version2 = "1.0.7";
 
    if (versionCompare(version1, version2) < 0)
      Console.WriteLine(version1 + " is smaller");
    else if (versionCompare(version1, version2) > 0)
      Console.WriteLine(version2 + " is smaller");
    else
      Console.WriteLine("Both version are equal");
  }
}
 
// This code is contributed by divyeshrabadiya07.

Javascript

// Javascript implementation
 
// Method to compare two versions.
// Returns 1 if v2 is smaller, -1
// if v1 is smaller, 0 if equal
function versionCompare(v1, v2)
{
    // vnum stores each numeric
    // part of version
    var vnum1 = 0, vnum2 = 0;
 
    // loop until both string are
    // processed
    for (var i = 0, j = 0; (i < v1.length
                            || j < v2.length);) {
        // storing numeric part of
        // version 1 in vnum1
        while (i < v1.length && v1[i] != '.') {
            vnum1 = vnum1 * 10 + (v1[i] - '0');
            i++;
        }
 
        // storing numeric part of
        // version 2 in vnum2
        while (j < v2.length && v2[j] != '.') {
            vnum2 = vnum2 * 10 + (v2[j] - '0');
            j++;
        }
 
        if (vnum1 > vnum2)
            return 1;
        if (vnum2 > vnum1)
            return -1;
 
        // if equal, reset variables and
        // go for next numeric part
        vnum1 = vnum2 = 0;
        i++;
        j++;
    }
    return 0;
}
 
// Driver method to check above
// comparison function
var version1 = "1.0.3";
var version2 = "1.0.7";
 
if (versionCompare(version1, version2) < 0)
    document.write(version1 + " is smaller" + "<br>");
else if (versionCompare(version1, version2) > 0)
    document.write(version2 + " is smaller" + "<br>");
else
    document.write("Both version are equal" + "<br>");
 
// This is code is contributed
// by shubhamsingh10

输出：

1.0.3 is greater

复杂性分析：

• 时间复杂度： O(n)，其中 n 是字符串的长度。
  只需要遍历一次字符串。
• 辅助空间： O(1)。
  因为不需要额外的空间。