给定一行一列中最大的元素时求原始矩阵
原文:https://www . geesforgeks . org/get-原始矩阵-当给定一行和一列中最大的元素时/
分别给定 N 和 M 整数的两个数组 A[] 和 B[] 。还给出了一个 N X M 二进制矩阵，其中 1 表示原矩阵中有一个正整数，0 表示原矩阵中该位置用 0 填充。任务是形成原始矩阵，使得 A[i] 表示第 i 第 行中的最大元素， B[j] 表示第 j 第列中的最大元素。 举例:****
Input: A[] = {2, 1, 3}, B[] = {2, 3, 0, 0, 2, 0, 1},
matrix[] = {{1, 0, 0, 0, 1, 0, 0},
{0, 0, 0, 0, 0, 0, 1},
{1, 1, 0, 0, 0, 0, 0}}
Output:
2 0 0 0 2 0 0
0 0 0 0 0 0 1
2 3 0 0 0 0 0
Input: A[] = {2, 4}, B[] = {4, 2},
matrix[] = {{1, 1},
{1, 1}}
Output:
2 2
4 2
方法:对矩阵中的每个指标 (i，j) 进行迭代，如果 mat[i][j] == 1 ，则用 min(A[i]，B[j]) 填充该位置。这是因为当前元素是第 i 第 行和第 j 第 列的一部分，如果选择了最大值(A[i]，B[j]) ，则无法满足其中一个条件，即所选元素可能超过当前行或当前列所需的最大元素。 以下是上述方法的实施:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define N 3
#define M 7
// Function that prints the original matrix
void printOriginalMatrix(int a[], int b[], int mat[N][M])
{
// Iterate in the row
for (int i = 0; i < N; i++) {
// Iterate in the column
for (int j = 0; j < M; j++) {
// If previously existed an element
if (mat[i][j] == 1)
cout << min(a[i], b[j]) << " ";
else
cout << 0 << " ";
}
cout << endl;
}
}
// Driver code
int main()
{
int a[] = { 2, 1, 3 };
int b[] = { 2, 3, 0, 0, 2, 0, 1 };
int mat[N][M] = { { 1, 0, 0, 0, 1, 0, 0 },
{ 0, 0, 0, 0, 0, 0, 1 },
{ 1, 1, 0, 0, 0, 0, 0 } };
printOriginalMatrix(a, b, mat);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
class GFG
{
static int N = 3;
static int M = 7;
// Function that prints the original matrix
static void printOriginalMatrix(int a[], int b[],
int[][] mat)
{
// Iterate in the row
for (int i = 0; i < N; i++)
{
// Iterate in the column
for (int j = 0; j < M; j++)
{
// If previously existed an element
if (mat[i][j] == 1)
System.out.print(Math.min(a[i],
b[j]) + " ");
else
System.out.print("0" + " ");
}
System.out.println();
}
}
// Driver code
public static void main(String[] args)
{
int a[] = { 2, 1, 3 };
int b[] = { 2, 3, 0, 0, 2, 0, 1 };
int[][] mat = {{ 1, 0, 0, 0, 1, 0, 0 },
{ 0, 0, 0, 0, 0, 0, 1 },
{ 1, 1, 0, 0, 0, 0, 0 }};
printOriginalMatrix(a, b, mat);
}
}
// This code is contributed by Code_Mech
Python 3
# Python3 implementation of the approach
N = 3
M = 7
# Function that prints the original matrix
def printOriginalMatrix(a, b, mat) :
# Iterate in the row
for i in range(N) :
# Iterate in the column
for j in range(M) :
# If previously existed an element
if (mat[i][j] == 1) :
print(min(a[i], b[j]), end = " ");
else :
print(0, end = " ");
print()
# Driver code
if __name__ == "__main__" :
a = [ 2, 1, 3 ]
b = [ 2, 3, 0, 0, 2, 0, 1 ]
mat = [[ 1, 0, 0, 0, 1, 0, 0 ],
[ 0, 0, 0, 0, 0, 0, 1 ],
[ 1, 1, 0, 0, 0, 0, 0 ]];
printOriginalMatrix(a, b, mat);
# This code is contributed by Ryuga
C
// C# implementation of the approach
using System;
class GFG
{
static int N = 3;
static int M = 7;
// Function that prints the original matrix
static void printOriginalMatrix(int[] a, int[] b,
int[,] mat)
{
// Iterate in the row
for (int i = 0; i < N; i++)
{
// Iterate in the column
for (int j = 0; j < M; j++)
{
// If previously existed an element
if (mat[i,j] == 1)
Console.Write(Math.Min(a[i],
b[j]) + " ");
else
Console.Write("0" + " ");
}
Console.WriteLine();
}
}
// Driver code
public static void Main()
{
int[] a = { 2, 1, 3 };
int[] b = { 2, 3, 0, 0, 2, 0, 1 };
int[,] mat = {{ 1, 0, 0, 0, 1, 0, 0 },
{ 0, 0, 0, 0, 0, 0, 1 },
{ 1, 1, 0, 0, 0, 0, 0 }};
printOriginalMatrix(a, b, mat);
}
}
// This code is contributed by Code_Mech
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of the approach
$N = 3;
$M = 7;
// Function that prints the original matrix
function printOriginalMatrix($a, $b, $mat)
{
// Iterate in the row
for ($i = 0; $i < $GLOBALS['N']; $i++)
{
// Iterate in the column
for ($j = 0; $j < $GLOBALS['M']; $j++)
{
// If previously existed an element
if ($mat[$i][$j] == 1)
echo min($a[$i], $b[$j])." ";
else
echo "0"." ";
}
echo "\r\n";
}
}
// Driver code
$a = array( 2, 1, 3 );
$b = array(2, 3, 0, 0, 2, 0, 1 );
$mat = array( array( 1, 0, 0, 0, 1, 0, 0 ),
array( 0, 0, 0, 0, 0, 0, 1 ),
array( 1, 1, 0, 0, 0, 0, 0 ));
printOriginalMatrix($a, $b, $mat);
// This code is contributed by Shashank_Sharma
?>
java 描述语言
<script>
// Javascript implementation of the approach
let N = 3;
let M = 7;
// Function that prints the original matrix
function printOriginalMatrix(a,b,mat)
{
// Iterate in the row
for (let i = 0; i < N; i++)
{
// Iterate in the column
for (let j = 0; j < M; j++)
{
// If previously existed an element
if (mat[i][j] == 1)
document.write(Math.min(a[i],
b[j]) + " ");
else
document.write("0" + " ");
}
document.write("<br>");
}
}
// Driver code
let a = [ 2, 1, 3 ];
let b = [ 2, 3, 0, 0, 2, 0, 1 ];
let mat = [[ 1, 0, 0, 0, 1, 0, 0 ],
[ 0, 0, 0, 0, 0, 0, 1 ],
[ 1, 1, 0, 0, 0, 0, 0 ]];
printOriginalMatrix(a, b, mat);
// This code is contributed Bobby
</script>
Output:
2 0 0 0 2 0 0
0 0 0 0 0 0 1
2 3 0 0 0 0 0
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