比较允许大输入的版本号
比较两个版本，version1，version2。
如果版本 1 > 版本 2 返回 1
如果版本 1 < 版本 2 返回 -1
如果版本 1 = 版本 2 返回 0
版本字符串是非空的，只包含数字和 '.'字符。这 '。'字符不代表小数点，用于分隔数字序列。
版本排序示例。
0.1 < 1.1 < 1.2 < 1.13 < 1.13.4
注意：这里字符串中的数字可能很大，所以不要尝试转换这些数字
到无符号长长。例如。版本1 = 1.234565434523423423523423423423434432.23.0
例子：
Input : version1 : 002.0005.12.3
version2 : 2.5.12.3
Output : 0
Input : version1 : 451231654684151546847799885544662
version2 : 1.256.24.5.5
Output : 1
Input : version1 : 1.21.20
version2 : 1.21.25
Output : -1
Input : version1 : 1.2
version2 : 1.2.0.0.0
Output : 0
Input : version1 : 1.2
version2 : 1.0.1
Output : -1我们在下面的帖子中讨论了一个解决方案。
比较两个版本号
上面讨论的之前的解决方案存在一些问题，例如，它不处理前导零并且不适用于大数字，因为版本号的各个部分存储为 int。
在该解决方案中，解决了上述问题。我们同时遍历这两个版本并处理它们，直到它们都被完全遍历。将版本 1 和版本 2 中的数字存储在不同的字符串中，即 substr_version1 和 substr_version2。比较这些子串，
如果 substr_version1 > substr_version2 的长度显然 substr_version1 的值更大，因此返回 +1。类似的情况是当 substr_version2 > substr_version1 时，我们将返回 -1。但是如果两个子串的长度相似，那么
我们必须检查两个子字符串中的每个字符，然后比较这些字符，然后适当地返回结果。
C++
// C++ program to compare two versions
#include
using namespace std;
// Utility function to compare each substring
// of version1 and version2
int compareSubstr(char* substr_version1,
char* substr_version2,
int len_substr_version1,
int len_substr_version2)
{
// If length of substring of version 1 is
// greater then it means value of substr
// of version1 is also greater
if (len_substr_version1 > len_substr_version2)
return 1;
else if (len_substr_version1 < len_substr_version2)
return -1;
// When length of the substrings of
// both versions is same.
else {
int i = 0, j = 0;
// Compare each character of both substrings
// and return accordingly.
while (i < len_substr_version1) {
if (substr_version1[i] < substr_version2[j])
return -1;
else if (substr_version1[i]
> substr_version2[j])
return 1;
i++, j++;
}
return 0;
}
}
// Function to compare two versions.
int compareVersion(char* version1, char* version2)
{
int len_version1 = strlen(version1);
int len_version2 = strlen(version2);
char* substr_version1
= (char*)malloc(sizeof(char) * 1000);
char* substr_version2
= (char*)malloc(sizeof(char) * 1000);
// Loop until both strings are exhausted.
// and extract the substrings from version1
// and version2
int i = 0, j = 0;
while (i < len_version1 || j < len_version2) {
int p = 0, q = 0;
// Skip the leading zeros in version1 string.
while (version1[i] == '0')
i++;
// Skip the leading zeros in version2 string.
while (version2[j] == '0')
j++;
// Extract the substring from version1.
while (version1[i] != '.' && i < len_version1)
substr_version1[p++] = version1[i++];
// Extract the substring from version2.
while (version2[j] != '.' && j < len_version2)
substr_version2[q++] = version2[j++];
int res = compareSubstr(substr_version1,
substr_version2, p, q);
// If res is either -1 or +1 then simply return.
if (res)
return res;
i++;
j++;
}
// Here both versions are exhausted it implicitly
// means that both strings are equal.
return 0;
}
// Driver code
int main()
{
// Define Two versions.
char version1[] = "1.2.032.45";
char version2[] = "1.2.32.4";
int res = compareVersion(version1, version2);
cout << res << endl;
return 0;
}
// This code is contributed by SHUBHAMSINGH10 C
/* C program to compare two versions */
#include
#include
#include
// utility function to compare each substring of version1
// and version2
int compareSubstr(char* substr_version1,
char* substr_version2,
int len_substr_version1,
int len_substr_version2)
{
// if length of substring of version 1 is greater then
// it means value of substr of version1 is also greater
if (len_substr_version1 > len_substr_version2)
return 1;
else if (len_substr_version1 < len_substr_version2)
return -1;
// when length of the substrings of both versions is
// same.
else {
int i = 0, j = 0;
// compare each character of both substrings and
// return accordingly.
while (i < len_substr_version1) {
if (substr_version1[i] < substr_version2[j])
return -1;
else if (substr_version1[i]
> substr_version2[j])
return 1;
i++, j++;
}
return 0;
}
}
// function to compare two versions.
int compareVersion(char* version1, char* version2)
{
int len_version1 = strlen(version1);
int len_version2 = strlen(version2);
char* substr_version1
= (char*)malloc(sizeof(char) * 1000);
char* substr_version2
= (char*)malloc(sizeof(char) * 1000);
// loop until both strings are exhausted.
// and extract the substrings from version1 and version2
int i = 0, j = 0;
while (i < len_version1 || j < len_version2) {
int p = 0, q = 0;
// skip the leading zeros in version1 string.
while (version1[i] == '0')
i++;
// skip the leading zeros in version2 string.
while (version2[j] == '0')
j++;
// extract the substring from version1.
while (version1[i] != '.' && i < len_version1)
substr_version1[p++] = version1[i++];
// extract the substring from version2.
while (version2[j] != '.' && j < len_version2)
substr_version2[q++] = version2[j++];
int res = compareSubstr(substr_version1,
substr_version2, p, q);
// if res is either -1 or +1 then simply return.
if (res)
return res;
i++;
j++;
}
// here both versions are exhausted it implicitly
// means that both strings are equal.
return 0;
}
// Driver code.
int main()
{
// Define Two versions.
char version1[] = "1.2.032.45";
char version2[] = "1.2.32.4";
int res = compareVersion(version1, version2);
printf("%d\n", res);
return 0;
} Java
// Java program to compare two versions
import java.io.*;
class GFG {
// utility function to compare each substring of
// version1 and version2
public static int compareSubstr(String substr_version1,
String substr_version2)
{
int len_substr_version1 = substr_version1.length();
int len_substr_version2 = substr_version2.length();
// If length of substring of version 1 is greater
// then it means value of substr of version1 is
// also greater
if (len_substr_version1 > len_substr_version2)
return 1;
else if (len_substr_version2 > len_substr_version1)
return -1;
// When length of the substrings of
// both versions is same.
int res
= substr_version1.compareTo(substr_version2);
if (res > 0)
return 1;
else if (res < 0)
return -1;
return 0;
}
// Function to compare two versions.
public static int compareVersion(String version1,
String version2)
{
String substr_version1[] = version1.split("[.]");
String substr_version2[] = version2.split("[.]");
int len_version1 = substr_version1.length;
int len_version2 = substr_version2.length;
int i = 0;
int j = 0;
// Loop until both strings are exhausted.
// and extract the substrings from version1
// and version2
while (i < len_version1 || j < len_version2) {
String x = "";
String y = "";
if (i < len_version1) {
// Skip the leading zeros in
// version1 string.
if (substr_version1[i].charAt(0) == '0') {
int len = substr_version1[i].length();
int k = 0;
while (k < len
&& substr_version1[i].charAt(k)
== '0') {
k++;
}
x += substr_version1[i].substring(k);
}
else
x += substr_version1[i];
}
if (j < len_version2) {
// Skip the leading zeros in version2
// string.
if (substr_version2[i].charAt(0) == '0') {
int len = substr_version2[i].length();
int k = 0;
while (k < len
&& substr_version2[i].charAt(k)
== '0') {
k++;
}
y += substr_version2[i].substring(k);
}
else
y = substr_version2[i];
}
// If res is either -1 or +1
// then simply return.
int res = compareSubstr(x, y);
if (res != 0)
return res;
i++;
j++;
}
// Here both versions are exhausted
// it implicitly means that both
// strings are equal.
return 0;
}
// Driver code.
public static void main(String[] args)
{
System.out.println(
compareVersion("1.2.032.45", "1.2.32.4"));
}
}
// This code is contributed by naresh_saharan151C#
// C# program to compare two versions
using System;
public class GFG {
// utility function to compare each Substring of
// version1 and version2
public static int compareSubstr(string substr_version1,
string substr_version2)
{
int len_substr_version1 = substr_version1.Length;
int len_substr_version2 = substr_version1.Length;
// If length of Substring of version 1 is greater
// then it means value of substr of version1 is
// also greater
if (len_substr_version1 > len_substr_version2)
return 1;
else if (len_substr_version2 < len_substr_version1)
return -1;
// When length of the Substrings of
// both versions is same.
int res
= substr_version1.CompareTo(substr_version2);
if (res > 0)
return -1;
else if (res < 0)
return 1;
return 0;
}
// Function to compare two versions.
public static int compareVersion(string version1,
string version2)
{
string[] substr_version1 = version1.Split("[.]");
string[] substr_version2 = version2.Split("[.]");
int len_version1 = substr_version1.Length;
int len_version2 = substr_version2.Length;
int i = 0;
int j = 0;
// Loop until both strings are exhausted.
// and extract the Substrings from version1
// and version2
while (i < len_version1 || j < len_version2) {
string x = "";
string y = "";
if (i < len_version1) {
// Skip the leading zeros in
// version1 string.
if (substr_version1[i][0] == '0') {
int len = substr_version1[i].Length;
int k = 0;
while (k < len
&& substr_version1[i][k]
== '0') {
k++;
}
x += substr_version1[i].Substring(k);
}
else
x += substr_version1[i];
}
if (j < len_version2) {
// Skip the leading zeros in version2
// string.
if (substr_version2[i][0] == '0') {
int len = substr_version2[i].Length;
int k = 0;
while (k < len
&& substr_version2[i][k]
== '0') {
k++;
}
y += substr_version2[i].Substring(k);
}
else
y = substr_version2[i];
}
// If res is either -1 or +1
// then simply return.
int res = compareSubstr(x, y);
if (res != 0)
return res;
i++;
j++;
}
// Here both versions are exhausted
// it implicitly means that both
// strings are equal.
return 0;
}
// Driver code.
static public void Main()
{
Console.Write(
compareVersion("1.2.032.45", "1.2.32.4"));
}
}
// This code is contributed by shubhamsingh10输出：
1时间复杂度： O(2 * N) –> O(N)
这里最坏的情况是两个版本相等，因此在提取两个子字符串后，它们中的每一个都将再次在 compareSubstr()函数中进行比较，这将使复杂度达到 (2 * N)。