与矩阵中的最大差异配对

给定一个 NxM 矩阵，其中 N 行 M 列正整数。任务是找到给定矩阵中具有最大差异的对。
注意：位置 (a, b) 和 (b, a) 处的对被认为是等效的。
例子：

Input : mat[N][M] = {{1, 2, 3, 4},
                 {25, 6, 7, 8},
                 {9, 10, 11, 12},
                 {13, 14, 15, 16}}
Output : 24
Pair (25, 1) has the maximum difference

Input : mat[N][M] = {{1, 2, 3},
                 {4, 6, 7},
                 {9, 10, 5}}
Output : 9
Pair (10, 1) has the maximum difference.

这个想法是观察对具有最大差异的对的贡献元素是矩阵中的最大和最小元素。因此，找到矩阵中的最大和最小元素并返回它们之间的差值。
下面是上述方法的实现：

C++

// C++ program to find with maximum
// difference in a matrix
 
#include 
using namespace std;
 
#define N 4 // Rows
#define M 4 // Columns
 
// Function to find pair with maximum
// difference in a matrix
int maxDifferencePair(int mat[N][M])
{
    int maxElement = INT_MIN; // max
    int minElement = INT_MAX; // min
 
    // Traverse the matrix
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < M; j++) {
            // Find max element
            if (mat[i][j] > maxElement) {
                maxElement = mat[i][j];
            }
 
            // Find min element
            if (mat[i][j] < minElement) {
                minElement = mat[i][j];
            }
        }
    }
 
    return abs(maxElement - minElement);
}
 
// Driver Code
int main()
{
    // matrix
    int mat[N][M] = { { 1, 2, 3, 4 },
                      { 25, 6, 7, 8 },
                      { 9, 10, 11, 12 },
                      { 13, 14, 15, 16 } };
 
    cout << maxDifferencePair(mat) << endl;
 
    return 0;
}

Java

// Java program to find with maximum
// difference in a matrix
 
import java.io.*;
 
class GFG {
  
static int N= 4; // Rows
static int  M = 4; // Columns
 
// Function to find pair with maximum
// difference in a matrix
static int maxDifferencePair(int mat[][])
{
    int maxElement = Integer.MIN_VALUE; // max
    int minElement = Integer.MAX_VALUE; // min
 
    // Traverse the matrix
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < M; j++) {
            // Find max element
            if (mat[i][j] > maxElement) {
                maxElement = mat[i][j];
            }
 
            // Find min element
            if (mat[i][j] < minElement) {
                minElement = mat[i][j];
            }
        }
    }
 
    return Math.abs(maxElement - minElement);
}
 
// Driver Code
 
    public static void main (String[] args) {
        // matrix
    int mat[][] = { { 1, 2, 3, 4 },
                    { 25, 6, 7, 8 },
                    { 9, 10, 11, 12 },
                    { 13, 14, 15, 16 } };
 
    System.out.println( maxDifferencePair(mat));
    }
}
 
// This code is contributed by inder_verma..

Python3

# Python3 program to find with maximum
# difference in a matrix
 
N = 4 # Rows
M = 4 # Columns
 
# Function to find pair with maximum
# difference in a matrix
def maxDifferencePair(mat):
 
    maxElement = -10**9 # max
    minElement = 10**9 # min
 
    # Traverse the matrix
    for i in range(N):
        for j in range(M):
             
            # Find max element
            if (mat[i][j] > maxElement):
                maxElement = mat[i][j]
 
            # Find min element
            if (mat[i][j] < minElement):
                minElement = mat[i][j]
 
    return abs(maxElement - minElement)
 
# Driver Code
 
# matrix
mat = [[ 1, 2, 3, 4 ],
       [ 25, 6, 7, 8 ],
       [ 9, 10, 11, 12 ],
       [ 13, 14, 15, 16]]
 
print(maxDifferencePair(mat))
 
# This code is contributed
# by mohit kumar

C#

// C# program to find with maximum
// difference in a matrix
using System;
 
class GFG
{
static int N = 4; // Rows
static int M = 4; // Columns
 
// Function to find pair with
// maximum difference in a matrix
static int maxDifferencePair(int [,]mat)
{
    int maxElement = int.MinValue; // max
    int minElement = int.MaxValue; // min
 
    // Traverse the matrix
    for (int i = 0; i < N; i++)
    {
        for (int j = 0; j < M; j++)
        {
            // Find max element
            if (mat[i, j] > maxElement)
            {
                maxElement = mat[i, j];
            }
 
            // Find min element
            if (mat[i, j] < minElement)
            {
                minElement = mat[i, j];
            }
        }
    }
 
    return Math.Abs(maxElement -
                    minElement);
}
 
// Driver Code
public static void Main ()
{
    // matrix
    int [,]mat = {{ 1, 2, 3, 4 },
                  { 25, 6, 7, 8 },
                  { 9, 10, 11, 12 },
                  { 13, 14, 15, 16 }};
     
    Console.WriteLine( maxDifferencePair(mat));
}
}
 
// This code is contributed
// by inder_verma

Javascript

 
// JavaScript program to find with maximum
// difference in a matrix
 
let N= 4; // Rows
let M = 4; // Columns
 
// Function to find pair with maximum
// difference in a matrix
function maxDifferencePair(mat)
{
    let maxElement = Number.MIN_VALUE; // max
    let minElement = Number.MAX_VALUE; // min
   
    // Traverse the matrix
    for (let i = 0; i < N; i++) {
        for (let j = 0; j < M; j++) {
            // Find max element
            if (mat[i][j] > maxElement) {
                maxElement = mat[i][j];
            }
   
            // Find min element
            if (mat[i][j] < minElement) {
                minElement = mat[i][j];
            }
        }
    }
   
    return Math.abs(maxElement - minElement);
}
 
// Driver Code
let mat = [[ 1, 2, 3, 4 ],
       [ 25, 6, 7, 8 ],
       [ 9, 10, 11, 12 ],
       [ 13, 14, 15, 16]];
        
document.write( maxDifferencePair(mat));
 
 
// This code is contributed by unknown2108
 

输出：

24

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