方阵中的最大值和最小值
原文： https://www.geeksforgeeks.org/maximum-and-minimum-in-a-square-matrix/
给定n * n阶方阵，从给定的矩阵中找到最大值和最小值。
示例：
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
朴素的方法：
我们使用线性搜索分别找到矩阵的最大值和最小值。 所需的比较数是n ^ 2用于找到最小值，n ^ 2用于找到最大值元素。 总比较等于2n ^ 2。
对比较（有效方法）：
从矩阵中选择两个元素，一个从矩阵行的开头选择，另一个从矩阵同一行的结尾选择，比较它们，然后将它们的较小值与矩阵的最小值比较小，将它们的较大值与矩阵的最大值比较。 我们可以看到，对于两个元素，我们需要 3 比较，因此对于遍历整个矩阵，我们总共需要3/2 n ^ 2比较。
注意：这是最大数组最小值的方法 3 的扩展形式。
C++
// C++ program for finding maximum and minimum in
// a matrix.
#include<bits/stdc++.h>
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 = +2147483647;
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 = +2147483647;
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
<?php
// PHP program for finding
// maximum and minimum in
// a matrix.
$MAX = 100;
// Finds maximum and minimum
// in arr[0..n-1][0..n-1]
// using pair wise comparisons
function maxMin($arr, $n)
{
$min = PHP_INT_MAX;
$max = PHP_INT_MIN;
// Traverses rows one by one
for ($i = 0; $i < $n; $i++)
{
for ($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];
}
}
}
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\.
?>
输出：
Maximum = 25, Minimum = 0
版权属于：月萌API www.moonapi.com，转载请注明出处
