本文概述
- 建议:在继续解决方案之前, 请先在{IDE}上尝试使用你的方法。
- C ++
- Java
- Python3
- C#
- 的PHP
例子:
Input: mat[4][4] = {{1, 2, 3, 4}, {25, 6, 7, 8}, {9, 10, 11, 12}, {13, 14, 15, 16}};
Output: 25Input: mat[3][4] = {{9, 8, 7, 6}, {5, 4, 3, 2}, {1, 0, 12, 45}};
Output: 45
推荐:请尝试以下方法{IDE}首先, 在继续解决方案之前。方法:这个想法是使用两个嵌套循环遍历矩阵, 一个嵌套行, 一个嵌套列, 并找到最大元素。使用最小值初始化变量maxElement并遍历矩阵, 并在每次比较当前元素大于maxElement时进行比较。如果是, 则使用当前元素更新maxElement。
下面是上述方法的实现:
C ++
// CPP code to find max element in a matrix
#include <
bits/stdc++.h>
using namespace std;
#define N 4
#define M 4// Function to find max element
// mat[][] : 2D array to find max element
int findMax( int mat[N][M])
{// Initializing max element as INT_MIN
int maxElement = INT_MIN;
// checking each element of matrix
// if it is greater than maxElement, // update maxElement
for ( int i = 0;
i <
N;
i++) {
for ( int j = 0;
j <
M;
j++) {
if (mat[i][j] >
maxElement) {
maxElement = mat[i][j];
}
}
}// finally return maxElement
return maxElement;
}// 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 <
<
findMax(mat) <
<
endl;
return 0;
}
Java
// Java code to find max element in a matrixpublic class GFG {final static int N = 4 ;
final static intM = 4 ;
// Function to find max element
// mat[][] : 2D array to find max element
static int findMax( int mat[][])
{// Initializing max element as INT_MIN
int maxElement = Integer.MIN_VALUE;
// checking each element of matrix
// if it is greater than maxElement, // update maxElement
for ( int i = 0 ;
i <
N;
i++) {
for ( int j = 0 ;
j <
M;
j++) {
if (mat[i][j] >
maxElement) {
maxElement = mat[i][j];
}
}
}// finally return maxElement
return maxElement;
}// 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(findMax(mat)) ;
}
// This Code is contributed by ANKITRAI1
}
Python3
# Python 3 code to find max element
# in a matrix
import sys
N = 4
M = 4# Function to find max element
# mat[][] : 2D array to find max element
def findMax(mat):# Initializing max element as INT_MIN
maxElement = - sys.maxsize - 1# checking each element of matrix
# if it is greater than maxElement, # update maxElement
for i in range (N):
for j in range (M):
if (mat[i][j] >
maxElement):
maxElement = mat[i][j]# finally return maxElement
return maxElement# Driver code
if __name__ = = '__main__' :# matrix
mat = [[ 1 , 2 , 3 , 4 ], [ 25 , 6 , 7 , 8 ], [ 9 , 10 , 11 , 12 ], [ 13 , 14 , 15 , 16 ]]
print (findMax(mat))# This code is contributed by
# Surendra_Gangwar
C#
// C# code to find max element in a matrix
using System;
class GFG {static int N = 4;
static int M = 4 ;
// Function to find max element
// mat[, ] : 2D array to find max element
static int findMax( int [, ] mat)
{// Initializing max element as INT_MIN
int maxElement = int .MinValue;
// checking each element of matrix
// if it is greater than maxElement, // update maxElement
for ( int i = 0;
i <
N;
i++) {for ( int j = 0;
j <
M;
j++) {if (mat[i, j] >
maxElement) {maxElement = mat[i, j];
}
}
}// finally return maxElement
return maxElement;
}// 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.Write(findMax(mat)) ;
}}// This code is contributed by ChitraNayal
的PHP
<
?php
// PHP code to find max element in a matrix// Function to find max element
// mat[][] : 2D array to find max element
function findMax( $mat )
{// Initializing max element as INT_MIN
$maxElement = PHP_INT_MIN;
// checking each element of matrix
// if it is greater than maxElement, // update maxElement
for ( $i = 0;
$i <
4;
$i ++)
{
for ( $j = 0;
$j <
4;
$j ++)
{
if ( $mat [ $i ][ $j ] >
$maxElement )
{
$maxElement = $mat [ $i ][ $j ];
}
}
}// finally return maxElement
return $maxElement ;
}// Driver code
$mat = array ( array (1, 2, 3, 4), array (25, 6, 7, 8), array (9, 10, 11, 12), array (13, 14, 15, 16));
echo findMax( $mat ) . "\n" ;
// This code is contributed
// by Akanksha Rai
?>
输出如下:
25
【算法设计(如何查找矩阵中的最大元素(详细实现))】时间复杂度:O(N * M)
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