方阵中的最大值和最小值。

给定一个 n*n 阶的方阵，从给定的矩阵中找出最大值和最小值。

例子：

Input : arr[][] = {5, 4, 9,
                   2, 0, 6,
                   3, 1, 8};
Output : Maximum = 9, Minimum = 0

Input : arr[][] = {-5, 3, 
                   2, 4};
Output : Maximum = 4, Minimum = -5

天真的方法：
我们使用线性搜索分别找到矩阵的最大值和最小值。需要比较的次数是 n ²用于查找最小值和 n ²用于查找最大元素。总比较等于 2n ² 。

配对比较（高效方法）：
从矩阵中选择两个元素，一个从矩阵行的开头，另一个从矩阵同一行的末尾，比较它们，然后将它们中的较小者与矩阵的最小值进行比较，将它们中的较大者与矩阵的最大值进行比较矩阵。我们可以看到，对于两个元素，我们需要 3 次比较，因此为了遍历整个矩阵，我们总共需要 3/2 n ^{2 次}比较。

注意：这是数组的最大最小值方法3的扩展形式。

C++

// C++ program for finding maximum and minimum in
// a matrix.
#include
using namespace std;
 
#define MAX 100
 
// Finds maximum and minimum in arr[0..n-1][0..n-1]
// using pair wise comparisons
void maxMin(int arr[][MAX], int n)
{
    int min = INT_MAX;
    int max = INT_MIN;
 
    // Traverses rows one by one
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j <= n/2; j++)
        {
            // Compare elements from beginning
            // and end of current row
            if (arr[i][j] > arr[i][n-j-1])
            {
                if (min > arr[i][n-j-1])
                    min = arr[i][n-j-1];
                if (max< arr[i][j])
                    max = arr[i][j];
            }
            else
            {
                if (min > arr[i][j])
                    min = arr[i][j];
                if (max< arr[i][n-j-1])
                    max = arr[i][n-j-1];
            }
        }
    }
    cout << "Maximum = " << max;
         << ", Minimum = " << min;
}
 
/* Driver program to test above function */
int main()
{
    int arr[MAX][MAX] = {5, 9, 11,
                        25, 0, 14,
                        21, 6, 4};
    maxMin(arr, 3);
    return 0;
}

Java

// Java program for finding maximum
// and minimum in a matrix.
 
class GFG
{
    static final int MAX = 100;
     
    // Finds maximum and minimum
    // in arr[0..n-1][0..n-1]
    // using pair wise comparisons
    static void maxMin(int arr[][], int n)
    {
        int min = [phone];
        int max = -2147483648;
     
        // Traverses rows one by one
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j <= n/2; j++)
            {
                // Compare elements from beginning
                // and end of current row
                if (arr[i][j] > arr[i][n - j - 1])
                {
                    if (min > arr[i][n - j - 1])
                        min = arr[i][n - j - 1];
                    if (max< arr[i][j])
                        max = arr[i][j];
                }
                else
                {
                    if (min > arr[i][j])
                        min = arr[i][j];
                    if (max< arr[i][n - j - 1])
                        max = arr[i][n - j - 1];
                }
            }
        }
        System.out.print("Maximum = "+max+
                         ", Minimum = "+min);
    }
     
    // Driver program
    public static void main (String[] args)
    {
        int arr[][] = {{5, 9, 11},
                       {25, 0, 14},
                       {21, 6, 4}};
        maxMin(arr, 3);
    }
}
 
// This code is contributed by Anant Agarwal.

Python3

# Python3 program for finding
# MAXimum and MINimum in a matrix.
MAX = 100
 
# Finds MAXimum and MINimum in arr[0..n-1][0..n-1]
# using pair wise comparisons
def MAXMIN(arr, n):
 
    MIN = 10**9
    MAX = -10**9
 
    # Traverses rows one by one
    for i in range(n):
        for j in range(n // 2 + 1):
             
        # Compare elements from beginning
        # and end of current row
            if (arr[i][j] > arr[i][n - j - 1]):
                if (MIN > arr[i][n - j - 1]):
                    MIN = arr[i][n - j - 1]
                if (MAX< arr[i][j]):
                    MAX = arr[i][j]
            else:
                if (MIN > arr[i][j]):
                    MIN = arr[i][j]
                if (MAX< arr[i][n - j - 1]):
                    MAX = arr[i][n - j - 1]
 
    print("MAXimum =", MAX, ", MINimum =", MIN)
 
# Driver Code
arr = [[5, 9, 11],
       [25, 0, 14],
       [21, 6, 4]]
 
MAXMIN(arr, 3)
 
# This code is contributed by Mohit Kumar

C#

// C# program for finding maximum
// and minimum in a matrix.
using System;
 
public class GFG {
     
    // Finds maximum and minimum
    // in arr[0..n-1][0..n-1]
    // using pair wise comparisons
    static void maxMin(int[,] arr, int n)
    {
        int min = [phone];
        int max = -2147483648;
     
        // Traverses rows one by one
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j <= n/2; j++)
            {
                 
                // Compare elements from beginning
                // and end of current row
                if (arr[i,j] > arr[i,n - j - 1])
                {
                    if (min > arr[i,n - j - 1])
                        min = arr[i,n - j - 1];
                    if (max < arr[i,j])
                        max = arr[i,j];
                }
                else
                {
                    if (min > arr[i,j])
                        min = arr[i,j];
                    if (max < arr[i,n - j - 1])
                        max = arr[i,n - j - 1];
                }
            }
        }
        Console.Write("Maximum = " + max +
                        ", Minimum = " + min);
    }
     
    // Driver code
    static public void Main ()
    {
        int[,] arr = { {5, 9, 11},
                       {25, 0, 14},
                       {21, 6, 4} };
                        
        maxMin(arr, 3);
    }
}
 
// This code is contributed by Shrikant13.

PHP

 $arr[$i][$n - $j - 1])
            {
                if ($min > $arr[$i][$n - $j - 1])
                    $min = $arr[$i][$n - $j - 1];
                if ($max< $arr[$i][$j])
                    $max = $arr[$i][$j];
            }
            else
            {
                if ($min > $arr[$i][$j])
                    $min = $arr[$i][$j];
                if ($max < $arr[$i][$n - $j - 1])
                    $max = $arr[$i][$n - $j - 1];
            }
        }
    }
    echo "Maximum = " , $max
        ,", Minimum = " , $min;
}
 
    // Driver Code
    $arr = array(array(5, 9, 11),
                array(25, 0, 14),
                array(21, 6, 4));
    maxMin($arr, 3);
     
// This code is contributed by anuj_67.
?>

Javascript

 
// Javascript program for finding maximum
// and minimum in a matrix.
let MAX = 100;
     
// Finds maximum and minimum
// in arr[0..n-1][0..n-1]
// using pair wise comparisons
function maxMin(arr,n)
{
    let min = [phone];
    let max = -2147483648;
 
    // Traverses rows one by one
    for(let i = 0; i < n; i++)
    {
        for(let j = 0; j <= n / 2; j++)
        {
             
            // Compare elements from beginning
            // and end of current row
            if (arr[i][j] > arr[i][n - j - 1])
            {
                if (min > arr[i][n - j - 1])
                    min = arr[i][n - j - 1];
                if (max< arr[i][j])
                    max = arr[i][j];
            }
            else
            {
                if (min > arr[i][j])
                    min = arr[i][j];
                if (max < arr[i][n - j - 1])
                    max = arr[i][n - j - 1];
            }
        }
    }
    document.write("Maximum = " + max +
                   ", Minimum = " + min);
}
 
// Driver Code
let arr = [ [ 5, 9, 11 ],
            [ 25, 0, 14 ],
            [ 21, 6, 4 ] ];
             
maxMin(arr, 3);
 
// This code is contributed by sravan kumar
 

输出：

Maximum = 25, Minimum = 0