从方阵对角线中找出最小和最大的元素
原文:https://www . geeksforgeeks . org/find-最小和最大元素-从正方形-矩阵-对角线/
给定一个 n*n 阶的方阵，从给定矩阵的两条对角线上找出最小和最大的元素。
示例:
Input : matrix = {
{1, 2, 3, 4, -10},
{5, 6, 7, 8, 6},
{1, 2, 11, 3, 4},
{5, 6, 70, 5, 8},
{4, 9, 7, 1, 5}};
Output :
Principal Diagonal Smallest Element: 1
Principal Diagonal Greatest Element :11
Secondary Diagonal Smallest Element: -10
Secondary Diagonal Greatest Element: 11
解决这个问题背后的思想是，首先检查遍历矩阵，并达到所有对角线元素(对于主对角线 i == j 和次对角线 i+j = size_of_matrix-1)，并将对角线元素与最小值和最大值变量进行比较，并获取新的最小值和最大值。二次对角线也是如此。
以下是上述方法的实现:
例 1:与 O(n^2)复杂性:
C++
// CPP program to find smallest and
// largest elements of both diagonals
#include<bits/stdc++.h>
using namespace std;
// Function to find smallest and largest element
// from principal and secondary diagonal
void diagonalsMinMax(int mat[5][5])
{
// take length of matrix
int n = sizeof(*mat) / 4;
if (n == 0)
return;
// declare and initialize variables
// with appropriate value
int principalMin = mat[0][0],
principalMax = mat[0][0];
int secondaryMin = mat[n - 1][0],
secondaryMax = mat[n - 1][0];
for (int i = 1; i < n; i++)
{
for (int j = 1; j < n; j++)
{
// Condition for principal
// diagonal
if (i == j)
{
// take new smallest value
if (mat[i][j] < principalMin)
{
principalMin = mat[i][j];
}
// take new largest value
if (mat[i][j] > principalMax)
{
principalMax = mat[i][j];
}
}
// Condition for secondary
// diagonal
if ((i + j) == (n - 1))
{
// take new smallest value
if (mat[i][j] < secondaryMin)
{
secondaryMin = mat[i][j];
}
// take new largest value
if (mat[i][j] > secondaryMax)
{
secondaryMax = mat[i][j];
}
}
}
}
cout << ("Principal Diagonal Smallest Element: ")
<< principalMin << endl;
cout << ("Principal Diagonal Greatest Element : ")
<< principalMax << endl;
cout << ("Secondary Diagonal Smallest Element: ")
<< secondaryMin << endl;
cout << ("Secondary Diagonal Greatest Element: ")
<< secondaryMax << endl;
}
// Driver code
int main()
{
// Declare and initialize 5X5 matrix
int matrix[5][5] = {{ 1, 2, 3, 4, -10 },
{ 5, 6, 7, 8, 6 },
{ 1, 2, 11, 3, 4 },
{ 5, 6, 70, 5, 8 },
{ 4, 9, 7, 1, -5 }};
diagonalsMinMax(matrix);
}
// This code is contributed by
// Shashank_Sharma
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to find
// smallest and largest elements of both diagonals
public class GFG {
// Function to find smallest and largest element from
// principal and secondary diagonal
static void diagonalsMinMax(int[][] mat)
{
// take length of matrix
int n = mat.length;
if (n == 0)
return;
// declare and initialize variables with appropriate value
int principalMin = mat[0][0], principalMax = mat[0][0];
int secondaryMin = mat[n-1][0], secondaryMax = mat[n-1][0];
for (int i = 1; i < n; i++) {
for (int j = 1; j < n; j++) {
// Condition for principal
// diagonal
if (i == j) {
// take new smallest value
if (mat[i][j] < principalMin) {
principalMin = mat[i][j];
}
// take new largest value
if (mat[i][j] > principalMax) {
principalMax = mat[i][j];
}
}
// Condition for secondary
// diagonal
if ((i + j) == (n - 1)) {
// take new smallest value
if (mat[i][j] < secondaryMin) {
secondaryMin = mat[i][j];
}
// take new largest value
if (mat[i][j] > secondaryMax) {
secondaryMax = mat[i][j];
}
}
}
}
System.out.println("Principal Diagonal Smallest Element: "
+ principalMin);
System.out.println("Principal Diagonal Greatest Element : "
+ principalMax);
System.out.println("Secondary Diagonal Smallest Element: "
+ secondaryMin);
System.out.println("Secondary Diagonal Greatest Element: "
+ secondaryMax);
}
// Driver code
static public void main(String[] args)
{
// Declare and initialize 5X5 matrix
int[][] matrix = {
{ 1, 2, 3, 4, -10 },
{ 5, 6, 7, 8, 6 },
{ 1, 2, 11, 3, 4 },
{ 5, 6, 70, 5, 8 },
{ 4, 9, 7, 1, -5 }
};
diagonalsMinMax(matrix);
}
}
Python 3
# Python3 program to find smallest and
# largest elements of both diagonals
# Function to find smallest and largest element
# from principal and secondary diagonal
def diagonalsMinMax(mat):
# take length of matrix
n = len(mat)
if (n == 0):
return
# declare and initialize variables
# with appropriate value
principalMin = mat[0][0]
principalMax = mat[0][0]
secondaryMin = mat[0][n-1]
secondaryMax = mat[0][n-1]
for i in range(1, n):
for j in range(1, n):
# Condition for principal
# diagonal
if (i == j):
# take new smallest value
if (mat[i][j] < principalMin):
principalMin = mat[i][j]
# take new largest value
if (mat[i][j] > principalMax):
principalMax = mat[i][j]
# Condition for secondary
# diagonal
if ((i + j) == (n - 1)):
# take new smallest value
if (mat[i][j] < secondaryMin):
secondaryMin = mat[i][j]
# take new largest value
if (mat[i][j] > secondaryMax):
secondaryMax = mat[i][j]
print("Principal Diagonal Smallest Element: ",
principalMin)
print("Principal Diagonal Greatest Element : ",
principalMax)
print("Secondary Diagonal Smallest Element: ",
secondaryMin)
print("Secondary Diagonal Greatest Element: ",
secondaryMax)
# Driver code
# Declare and initialize 5X5 matrix
matrix = [[ 1, 2, 3, 4, -10 ],
[ 5, 6, 7, 8, 6 ],
[ 1, 2, 11, 3, 4 ],
[ 5, 6, 70, 5, 8 ],
[ 4, 9, 7, 1, -5 ]]
diagonalsMinMax(matrix)
# This code is contributed by Mohit kumar 29
C
// C# program to find smallest and largest
// elements of both diagonals
using System;
public class GFG {
// Function to find smallest and largest element from
// principal and secondary diagonal
static void diagonalsMinMax(int[,] mat)
{
// take length of square matrix
int n = mat.GetLength(0);
if (n == 0)
return;
// declare and initialize variables with appropriate value
int principalMin = mat[0,0], principalMax = mat[0,0];
int secondaryMin = mat[n-1,0], secondaryMax = mat[n-1,0];
for (int i = 1; i < n; i++) {
for (int j = 1; j < n; j++) {
// Condition for principal
// diagonal
if (i == j) {
// take new smallest value
if (mat[i,j] < principalMin) {
principalMin = mat[i,j];
}
// take new largest value
if (mat[i,j] > principalMax) {
principalMax = mat[i,j];
}
}
// Condition for secondary
// diagonal
if ((i + j) == (n - 1)) {
// take new smallest value
if (mat[i,j] < secondaryMin) {
secondaryMin = mat[i,j];
}
// take new largest value
if (mat[i,j] > secondaryMax) {
secondaryMax = mat[i,j];
}
}
}
}
Console.WriteLine("Principal Diagonal Smallest Element: "
+ principalMin);
Console.WriteLine("Principal Diagonal Greatest Element : "
+ principalMax);
Console.WriteLine("Secondary Diagonal Smallest Element: "
+ secondaryMin);
Console.WriteLine("Secondary Diagonal Greatest Element: "
+ secondaryMax);
}
// Driver code
static void Main()
{
// Declare and initialize 5X5 matrix
int[,] matrix = {
{ 1, 2, 3, 4, -10 },
{ 5, 6, 7, 8, 6 },
{ 1, 2, 11, 3, 4 },
{ 5, 6, 70, 5, 8 },
{ 4, 9, 7, 1, -5 }
};
diagonalsMinMax(matrix);
}
// This code is contributed by Ryuga
}
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to find smallest and
// largest elements of both diagonals
// Function to find smallest and largest element
// from principal and secondary diagonal
function diagonalsMinMax($mat)
{
// take length of $matrix
$n = count($mat);
if ($n == 0)
return;
// declare and initialize variables
// with appropriate value
$principalMin = $mat[0][0];
$principalMax = $mat[0][0];
$secondaryMin = $mat[$n - 1][0];
$secondaryMax = $mat[$n - 1][0];
for ($i = 1; $i < $n; $i++)
{
for ($j = 1; $j < $n; $j++)
{
// Condition for principal
// diagonal
if ($i == $j)
{
// take new smallest value
if ($mat[$i][$j] < $principalMin)
{
$principalMin = $mat[$i][$j];
}
// take new largest value
if ($mat[$i][$j] > $principalMax)
{
$principalMax = $mat[$i][$j];
}
}
// Condition for secondary
// diagonal
if (($i + $j) == ($n - 1))
{
// take new smallest value
if ($mat[$i][$j] < $secondaryMin)
{
$secondaryMin = $mat[$i][$j];
}
// take new largest value
if ($mat[$i][$j] > $secondaryMax)
{
$secondaryMax = $mat[$i][$j];
}
}
}
}
echo "Principal Diagonal Smallest Element: ",
$principalMin, "\n";
echo "Principal Diagonal Greatest Element : ",
$principalMax, "\n";
echo "Secondary Diagonal Smallest Element: ",
$secondaryMin, "\n";
echo "Secondary Diagonal Greatest Element: ",
$secondaryMax, "\n";
}
// Driver code
// Declare and initialize 5X5 matrix
$matrix = array(array ( 1, 2, 3, 4, -10 ),
array ( 5, 6, 7, 8, 6 ),
array ( 1, 2, 11, 3, 4 ),
array ( 5, 6, 70, 5, 8 ),
array ( 4, 9, 7, 1, -5 ));
diagonalsMinMax($matrix);
// This code is contributed by
// ihritik
?>
java 描述语言
<script>
// Javascript program to find smallest and
// largest elements of both diagonals
// Function to find smallest and largest element
// from principal and secondary diagonal
function diagonalsMinMax(mat)
{
// take length of matrix
let n = mat.length;
if (n == 0)
return;
// declare and initialize variables
// with appropriate value
let principalMin = mat[0][0],
principalMax = mat[0][0];
let secondaryMin = mat[n - 1][0],
secondaryMax = mat[n - 1][0];
for (let i = 1; i < n; i++)
{
for (let j = 1; j < n; j++)
{
// Condition for principal
// diagonal
if (i == j)
{
// take new smallest value
if (mat[i][j] < principalMin)
{
principalMin = mat[i][j];
}
// take new largest value
if (mat[i][j] > principalMax)
{
principalMax = mat[i][j];
}
}
// Condition for secondary
// diagonal
if ((i + j) == (n - 1))
{
// take new smallest value
if (mat[i][j] < secondaryMin)
{
secondaryMin = mat[i][j];
}
// take new largest value
if (mat[i][j] > secondaryMax)
{
secondaryMax = mat[i][j];
}
}
}
}
document.write("Principal Diagonal Smallest Element: "
+ principalMin + "<br>");
document.write("Principal Diagonal Greatest Element : "
+ principalMax + "<br>");
document.write("Secondary Diagonal Smallest Element: "
+ secondaryMin + "<br>");
document.write("Secondary Diagonal Greatest Element: "
+ secondaryMax + "<br>");
}
// Driver code
// Declare and initialize 5X5 matrix
let matrix = [[ 1, 2, 3, 4, -10 ],
[ 5, 6, 7, 8, 6 ],
[ 1, 2, 11, 3, 4 ],
[ 5, 6, 70, 5, 8 ],
[ 4, 9, 7, 1, -5 ]];
diagonalsMinMax(matrix);
// This code is contributed by subham348.
</script>
Output:
Principal Diagonal Smallest Element: -5
Principal Diagonal Greatest Element : 11
Secondary Diagonal Smallest Element: 4
Secondary Diagonal Greatest Element: 11
示例 2:具有 O(n)复杂度:
C++
// C++ program to find
// smallest and largest elements of both diagonals
#include<bits/stdc++.h>
using namespace std;
const int n = 5;
// Function to find smallest and largest element
// from principal and secondary diagonal
void diagonalsMinMax(int mat [n][n])
{
// take length of matrix
if (n == 0)
return;
// declare and initialize variables
// with appropriate value
int principalMin = mat[0][0],
principalMax = mat[0][0];
int secondaryMin = mat[n - 1][0],
secondaryMax = mat[n - 1][0];
for (int i = 0; i < n; i++)
{
// Condition for principal
// diagonal mat[i][i]
// take new smallest value
if (mat[i][i] < principalMin)
{
principalMin = mat[i][i];
}
// take new largest value
if (mat[i][i] > principalMax)
{
principalMax = mat[i][i];
}
// Condition for secondary
// diagonal is mat[n-1-i][i]
// take new smallest value
if (mat[n - 1 - i][i] < secondaryMin)
{
secondaryMin = mat[n - 1 - i][i];
}
// take new largest value
if (mat[n - 1 - i][i] > secondaryMax)
{
secondaryMax = mat[n - 1 - i][i];
}
}
cout << "Principal Diagonal Smallest Element: "
<< principalMin << "\n";
cout << "Principal Diagonal Greatest Element : "
<< principalMax << "\n";
cout << "Secondary Diagonal Smallest Element: "
<< secondaryMin << "\n";
cout << "Secondary Diagonal Greatest Element: "
<< secondaryMax;
}
// Driver code
int main()
{
// Declare and initialize 5X5 matrix
int matrix [n][n] = {{ 1, 2, 3, 4, -10 },
{ 5, 6, 7, 8, 6 },
{ 1, 2, 11, 3, 4 },
{ 5, 6, 70, 5, 8 },
{ 4, 9, 7, 1, -5 }};
diagonalsMinMax(matrix);
}
// This code is contributed by ihritik
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to find
// smallest and largest elements of both diagonals
public class GFG {
// Function to find smallest and largest element from
// principal and secondary diagonal
static void diagonalsMinMax(int[][] mat)
{
// take length of matrix
int n = mat.length;
if (n == 0)
return;
// declare and initialism variables with appropriate value
int principalMin = mat[0][0], principalMax = mat[0][0];
int secondaryMin = mat[n-1][0], secondaryMax = mat[n-1][0];
for (int i = 0; i < n; i++) {
// Condition for principal
// diagonal mat[i][i]
// take new smallest value
if (mat[i][i] < principalMin) {
principalMin = mat[i][i];
}
// take new largest value
if (mat[i][i] > principalMax) {
principalMax = mat[i][i];
}
// Condition for secondary
// diagonal is mat[n-1-i][i]
// take new smallest value
if (mat[n - 1 - i][i] < secondaryMin) {
secondaryMin = mat[n - 1 - i][i];
}
// take new largest value
if (mat[n - 1 - i][i] > secondaryMax) {
secondaryMax = mat[n - 1 - i][i];
}
}
System.out.println("Principal Diagonal Smallest Element: "
+ principalMin);
System.out.println("Principal Diagonal Greatest Element : "
+ principalMax);
System.out.println("Secondary Diagonal Smallest Element: "
+ secondaryMin);
System.out.println("Secondary Diagonal Greatest Element: "
+ secondaryMax);
}
// Driver code
static public void main(String[] args)
{
// Declare and initialize 5X5 matrix
int[][] matrix = {
{ 1, 2, 3, 4, -10 },
{ 5, 6, 7, 8, 6 },
{ 1, 2, 11, 3, 4 },
{ 5, 6, 70, 5, 8 },
{ 4, 9, 7, 1, -5 }
};
diagonalsMinMax(matrix);
}
}
计算机编程语言
# Python3 program to find smallest and
# largest elements of both diagonals
n = 5
# Function to find smallest and largest element
# from principal and secondary diagonal
def diagonalsMinMax(mat):
# take length of matrix
if (n == 0):
return
# declare and initialize variables
# with appropriate value
principalMin = mat[0][0]
principalMax = mat[0][0]
secondaryMin = mat[n - 1][0]
secondaryMax = mat[n - 1][0]
for i in range(n):
# Condition for principal
# diagonal mat[i][i]
# take new smallest value
if (mat[i][i] < principalMin):
principalMin = mat[i][i]
# take new largest value
if (mat[i][i] > principalMax):
principalMax = mat[i][i]
# Condition for secondary
# diagonal is mat[n-1-i][i]
# take new smallest value
if (mat[n - 1 - i][i] < secondaryMin):
secondaryMin = mat[n - 1 - i][i]
# take new largest value
if (mat[n - 1 - i][i] > secondaryMax):
secondaryMax = mat[n - 1 - i][i]
print("Principal Diagonal Smallest Element: ",principalMin)
print("Principal Diagonal Greatest Element : ",principalMax)
print("Secondary Diagonal Smallest Element: ",secondaryMin)
print("Secondary Diagonal Greatest Element: ",secondaryMax)
# Driver code
# Declare and initialize 5X5 matrix
matrix= [[ 1, 2, 3, 4, -10 ],
[ 5, 6, 7, 8, 6 ],
[ 1, 2, 11, 3, 4 ],
[ 5, 6, 70, 5, 8 ],
[ 4, 9, 7, 1, -5 ]]
diagonalsMinMax(matrix)
# This code is contributed by mohit kumar 29
C
// C# program to find smallest and largest
// elements of both diagonals
using System;
public class GFG {
// Function to find smallest and largest element from
// principal and secondary diagonal
static void diagonalsMinMax(int[,] mat)
{
// take length of square matrix
int n = mat.GetLength(0);
if (n == 0)
return;
// declare and initialize variables with appropriate value
int principalMin = mat[0,0], principalMax = mat[0,0];
int secondaryMin = mat[n-1,0], secondaryMax = mat[n-1,0];
for (int i = 0; i < n; i++) {
// Condition for principal
// diagonal mat[i][i]
// take new smallest value
if (mat[i,i] < principalMin) {
principalMin = mat[i,i];
}
// take new largest value
if (mat[i,i] > principalMax) {
principalMax = mat[i,i];
}
// Condition for secondary
// diagonal is mat[n-1-i][i]
// take new smallest value
if (mat[n - 1 - i,i] < secondaryMin) {
secondaryMin = mat[n - 1 - i,i];
}
// take new largest value
if (mat[n - 1 - i,i] > secondaryMax) {
secondaryMax = mat[n - 1 - i,i];
}
}
Console.WriteLine("Principal Diagonal Smallest Element: "
+ principalMin);
Console.WriteLine("Principal Diagonal Greatest Element : "
+ principalMax);
Console.WriteLine("Secondary Diagonal Smallest Element: "
+ secondaryMin);
Console.WriteLine("Secondary Diagonal Greatest Element: "
+ secondaryMax);
}
// Driver code
public static void Main()
{
// Declare and initialize 5X5 matrix
int[,] matrix = {
{ 1, 2, 3, 4, -10 },
{ 5, 6, 7, 8, 6 },
{ 1, 2, 11, 3, 4 },
{ 5, 6, 70, 5, 8 },
{ 4, 9, 7, 1, -5 }
};
diagonalsMinMax(matrix);
}
}
/*This code is contributed by 29AjayKumar*/
java 描述语言
<script>
// JavaScript program to find
// smallest and largest elements
// of both diagonals
// Function to find smallest
// and largest element from
// principal and secondary diagonal
function diagonalsMinMax(mat)
{
// take length of matrix
let n = mat.length;
if (n == 0)
return;
// declare and initialism variables
// with appropriate value
let principalMin = mat[0][0],
principalMax = mat[0][0];
let secondaryMin = mat[n-1][0],
secondaryMax = mat[n-1][0];
for (let i = 0; i < n; i++) {
// Condition for principal
// diagonal mat[i][i]
// take new smallest value
if (mat[i][i] < principalMin) {
principalMin = mat[i][i];
}
// take new largest value
if (mat[i][i] > principalMax) {
principalMax = mat[i][i];
}
// Condition for secondary
// diagonal is mat[n-1-i][i]
// take new smallest value
if (mat[n - 1 - i][i] < secondaryMin) {
secondaryMin = mat[n - 1 - i][i];
}
// take new largest value
if (mat[n - 1 - i][i] > secondaryMax) {
secondaryMax = mat[n - 1 - i][i];
}
}
document.write("Principal Diagonal Smallest Element: "
+ principalMin+"<br>");
document.write("Principal Diagonal Greatest Element : "
+ principalMax+"<br>");
document.write("Secondary Diagonal Smallest Element: "
+ secondaryMin+"<br>");
document.write("Secondary Diagonal Greatest Element: "
+ secondaryMax+"<br>");
}
// Driver code
// Declare and initialize 5X5 matrix
let matrix = [
[ 1, 2, 3, 4, -10 ],
[ 5, 6, 7, 8, 6 ],
[ 1, 2, 11, 3, 4 ],
[ 5, 6, 70, 5, 8 ],
[ 4, 9, 7, 1, -5 ]
];
diagonalsMinMax(matrix);
// This code is contributed by sravan kumar
</script>
Output:
Principal Diagonal Smallest Element: -5
Principal Diagonal Greatest Element : 11
Secondary Diagonal Smallest Element: -10
Secondary Diagonal Greatest Element: 11
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