程序查找任意两个不同数字的索引之间的最大差

给定一个由N个整数组成的数组。任务是找到任意两个不同数字的索引之间的最大差。请注意，至少有两个不同的数字。

例子：

Input: a[] = {1, 2, 3, 2, 3}
Output: 4
The difference between 1 and last 3.

Input: a[] = {1, 1, 3, 1, 1, 1}
Output: 3
The difference between the index of 3 and last 1.

方法：首先，从头开始检查与a [0]不同的第一个数字，并将其索引的差存储为ind1 。另外，从头开始检查与a [n – 1]不同的第一个数字，并将其索引的差存储为ind2 。答案将是max(ind1，ind2) 。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
// Function to return the maximum difference
int findMaximumDiff(int a[], int n)
{
    int ind1 = 0;
  
    // Iteratively check from back
    for (int i = n - 1; i > 0; i--) {
  
        // Different numbers
        if (a[0] != a[i]) {
            ind1 = i;
            break;
        }
    }
  
    int ind2 = 0;
  
    // Iteratively check from
    // the beginning
    for (int i = 0; i < n - 1; i++) {
  
        // Different numbers
        if (a[n - 1] != a[i]) {
            ind2 = (n - 1 - i);
            break;
        }
    }
  
    return max(ind1, ind2);
}
  
// Driver code
int main()
{
    int a[] = { 1, 2, 3, 2, 3 };
    int n = sizeof(a) / sizeof(a[0]);
    cout << findMaximumDiff(a, n);
  
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
      
// Function to return the maximum difference
static int findMaximumDiff(int []a, int n)
{
    int ind1 = 0;
  
    // Iteratively check from back
    for (int i = n - 1; i > 0; i--) 
    {
  
        // Different numbers
        if (a[0] != a[i]) 
        {
            ind1 = i;
            break;
        }
    }
  
    int ind2 = 0;
  
    // Iteratively check from
    // the beginning
    for (int i = 0; i < n - 1; i++)
    {
  
        // Different numbers
        if (a[n - 1] != a[i]) 
        {
            ind2 = (n - 1 - i);
            break;
        }
    }
  
    return Math.max(ind1, ind2);
}
  
// Driver code
public static void main(String args[])
{
    int []a = { 1, 2, 3, 2, 3 };
    int n = a.length;
    System.out.println(findMaximumDiff(a, n));
}
}
  
// This code is contributed by Akanksha_Rai

Python3

# Python3 implementation of the approach
  
# Function to return the maximum difference
def findMaximumDiff(a, n):
    ind1 = 0
  
    # Iteratively check from back
    for i in range(n - 1, -1, -1):
  
        # Different numbers
        if (a[0] != a[i]):
            ind1 = i
            break
  
    ind2 = 0
  
    # Iteratively check from
    # the beginning
    for i in range(n - 1):
  
        # Different numbers
        if (a[n - 1] != a[i]):
            ind2 = (n - 1 - i)
            break
  
    return max(ind1, ind2)
  
# Driver code
a = [1, 2, 3, 2, 3]
n = len(a)
print(findMaximumDiff(a, n))
  
# This code is contributed by mohit kumar

C#

// C# implementation of the approach
using System;
  
class GFG
{
      
// Function to return the maximum difference
static int findMaximumDiff(int []a, int n)
{
    int ind1 = 0;
  
    // Iteratively check from back
    for (int i = n - 1; i > 0; i--) 
    {
  
        // Different numbers
        if (a[0] != a[i]) 
        {
            ind1 = i;
            break;
        }
    }
  
    int ind2 = 0;
  
    // Iteratively check from
    // the beginning
    for (int i = 0; i < n - 1; i++)
    {
  
        // Different numbers
        if (a[n - 1] != a[i]) 
        {
            ind2 = (n - 1 - i);
            break;
        }
    }
  
    return Math.Max(ind1, ind2);
}
  
// Driver code
static void Main()
{
    int []a = { 1, 2, 3, 2, 3 };
    int n = a.Length;
    Console.WriteLine(findMaximumDiff(a, n));
}
}
  
// This code is contributed by mits

PHP

 0; $i--) 
    { 
  
        // Different numbers 
        if ($a[0] != $a[$i])
        { 
            $ind1 = $i; 
            break; 
        } 
    } 
  
    $ind2 = 0; 
  
    // Iteratively check from 
    // the beginning 
    for ($i = 0; $i < $n - 1; $i++) 
    { 
  
        // Different numbers 
        if ($a[$n - 1] != $a[$i]) 
        { 
            $ind2 = ($n - 1 - $i); 
            break; 
        } 
    } 
  
    return max($ind1, $ind2); 
} 
  
// Driver code 
$a = array( 1, 2, 3, 2, 3 ); 
$n = count($a);
  
echo findMaximumDiff($a, $n); 
  
// This code is contributed by Ryuga
?>

输出：

4

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