给定两个整数的NXM矩阵。在一个运算中，我们可以转置矩阵1中的任何正方形子矩阵。任务是检查是否可以通过给定的操作将matrix1转换为matrix2。
例子：
Input: matrix1[][] = {
{1, 2, 3},
{4, 5, 6},
{7, 8, 9}},
matrix2[][] = {
{1, 4, 7},
{2, 5, 6},
{3, 8, 9}}
Output: Yes
Transpose the whole matrix and then we transpose the
sub-matrix with corners in cells (2, 2) and (3, 3).
Input: matrix1[][] = {
{1, 2},
{3, 4}},
matrix2[][] = {
{1, 4},
{3, 8}}
Output: No
方法：对两个矩阵的对角线进行排序并进行比较。如果对角线排序后两个矩阵相等，则可以将matrix1转换为matrix2。下面的方法是正确的，因为在转置数字时只能将数字转置为其对角线中的任何一个。
下面是上述方法的实现。
C++
// C++ implementation of the approach
#include
using namespace std;
#define n 3
#define m 3
// Function that returns true if matrix1
// can be converted to matrix2
// with the given operation
bool check(int a[n][m], int b[n][m])
{
// Traverse all the diagonals
// starting at first column
for (int i = 0; i < n; i++) {
vector v1, v2;
int r = i;
int col = 0;
// Traverse in diagonal
while (r >= 0 && col < m) {
// Store the diagonal elements
v1.push_back(a[r][col]);
v2.push_back(b[r][col]);
// Move up
r--;
col++;
}
// Sort the elements
sort(v1.begin(), v1.end());
sort(v2.begin(), v2.end());
// Check if they are same
for (int i = 0; i < v1.size(); i++) {
if (v1[i] != v2[i])
return false;
}
}
// Traverse all the diagonals
// starting at last row
for (int j = 1; j < m; j++) {
vector v1, v2;
int r = n - 1;
int col = j;
// Traverse in the diagonal
while (r >= 0 && col < m) {
// Store diagonal elements
v1.push_back(a[r][col]);
v2.push_back(b[r][col]);
r--;
col++;
}
// Sort all elements
sort(v1.begin(), v1.end());
sort(v2.begin(), v2.end());
// Check for same
for (int i = 0; i < v1.size(); i++) {
if (v1[i] != v2[i])
return false;
}
}
// If every element matches
return true;
}
// Driver code
int main()
{
int a[n][m] = { { 1, 2, 3 }, { 4, 5, 6 }, { 7, 8, 9 } };
int b[n][m] = { { 1, 4, 7 }, { 2, 5, 6 }, { 3, 8, 9 } };
if (check(a, b))
cout << "Yes";
else
cout << "No";
return 0;
} Java
// Java implementation of the approach
import java.util.*;
class GFG
{
static final int n = 3;
static final int m = 3;
// Function that returns true if matrix1
// can be converted to matrix2
// with the given operation
static boolean check(int a[][], int b[][])
{
// Traverse all the diagonals
// starting at first column
for (int i = 0; i < n; i++)
{
Vector v1 = new Vector();
Vector v2 = new Vector();
int r = i;
int col = 0;
// Traverse in diagonal
while (r >= 0 && col < m)
{
// Store the diagonal elements
v1.add(a[r][col]);
v2.add(b[r][col]);
// Move up
r--;
col++;
}
// Sort the elements
Collections.sort(v1);
Collections.sort(v2);
// Check if they are same
for (int j = 0; j < v1.size(); j++)
{
if (v1.get(j) != v2.get(j))
{
return false;
}
}
}
// Traverse all the diagonals
// starting at last row
for (int j = 1; j < m; j++)
{
Vector v1 = new Vector();
Vector v2 = new Vector();
int r = n - 1;
int col = j;
// Traverse in the diagonal
while (r >= 0 && col < m)
{
// Store diagonal elements
v1.add(a[r][col]);
v2.add(b[r][col]);
r--;
col++;
}
// Sort all elements
Collections.sort(v1);
Collections.sort(v2);
// Check for same
for (int i = 0; i < v1.size(); i++)
{
if (v1.get(i) != v2.get(i))
{
return false;
}
}
}
// If every element matches
return true;
}
// Driver code
public static void main(String[] args)
{
int a[][] = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}};
int b[][] = {{1, 4, 7}, {2, 5, 6}, {3, 8, 9}};
if (check(a, b))
{
System.out.println("Yes");
}
else
{
System.out.println("No");
}
}
}
// This code contributed by Rajput-Ji Python3
# Python 3 implementation of the approach
n = 3
m = 3
# Function that returns true if matrix1
# can be converted to matrix2
# with the given operation
def check(a, b):
# Traverse all the diagonals
# starting at first column
for i in range(n):
v1 = []
v2 = []
r = i
col = 0
# Traverse in diagonal
while (r >= 0 and col < m):
# Store the diagonal elements
v1.append(a[r][col])
v2.append(b[r][col])
# Move up
r -= 1
col += 1
# Sort the elements
v1.sort(reverse = False)
v2.sort(reverse = False)
# Check if they are same
for i in range(len(v1)):
if (v1[i] != v2[i]):
return False
# Traverse all the diagonals
# starting at last row
for j in range(1, m):
v1 = []
v2 = []
r = n - 1
col = j
# Traverse in the diagonal
while (r >= 0 and col < m):
# Store diagonal elements
v1.append(a[r][col])
v2.append(b[r][col])
r -= 1
col += 1
# Sort all elements
v1.sort(reverse = False)
v2.sort(reverse = False)
# Check for same
for i in range(len(v1)):
if (v1[i] != v2[i]):
return False
# If every element matches
return True
# Driver code
if __name__ == '__main__':
a = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
b = [[1, 4, 7], [2, 5, 6], [3, 8, 9]]
if (check(a, b)):
print("Yes")
else:
print("No")
# This code is contributed by
# Surendra_GangwarC#
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
static readonly int n = 3;
static readonly int m = 3;
// Function that returns true if matrix1
// can be converted to matrix2
// with the given operation
static bool check(int [,]a, int [,]b)
{
// Traverse all the diagonals
// starting at first column
for (int i = 0; i < n; i++)
{
List v1 = new List();
List v2 = new List();
int r = i;
int col = 0;
// Traverse in diagonal
while (r >= 0 && col < m)
{
// Store the diagonal elements
v1.Add(a[r, col]);
v2.Add(b[r, col]);
// Move up
r--;
col++;
}
// Sort the elements
v1.Sort();
v2.Sort();
// Check if they are same
for (int j = 0; j < v1.Count; j++)
{
if (v1[j] != v2[j])
{
return false;
}
}
}
// Traverse all the diagonals
// starting at last row
for (int j = 1; j < m; j++)
{
List v1 = new List();
List v2 = new List();
int r = n - 1;
int col = j;
// Traverse in the diagonal
while (r >= 0 && col < m)
{
// Store diagonal elements
v1.Add(a[r, col]);
v2.Add(b[r, col]);
r--;
col++;
}
// Sort all elements
v1.Sort();
v2.Sort();
// Check for same
for (int i = 0; i < v1.Count; i++)
{
if (v1[i] != v2[i])
{
return false;
}
}
}
// If every element matches
return true;
}
// Driver code
public static void Main()
{
int [,]a = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}};
int [,]b = {{1, 4, 7}, {2, 5, 6}, {3, 8, 9}};
if (check(a, b))
{
Console.WriteLine("Yes");
}
else
{
Console.WriteLine("No");
}
}
}
/* This code contributed by PrinciRaj1992 */ PHP
= 0 && $col < $m)
{
// Store the diagonal elements
array_push($v1, $a[$r][$col]);
array_push($v2, $b[$r][$col]);
// Move up
$r--;
$col++;
}
// Sort the elements
sort($v1);
sort($v2);
// Check if they are same
for ($i = 0; $i < count($v1); $i++)
{
if ($v1[$i] != $v2[$i])
return false;
}
}
// Traverse all the diagonals
// starting at last row
for ($j = 1; $j < $m; $j++)
{
$v1 = array();
$v2 = array();
$r = $n - 1;
$col = $j;
// Traverse in the diagonal
while ($r >= 0 && $col < $m)
{
// Store diagonal elements
array_push($v1, $a[$r][$col]);
array_push($v2, $b[$r][$col]);
$r--;
$col++;
}
// Sort all elements
sort($v1);
sort($v2);
// Check for same
for ($i = 0; $i < count($v1); $i++)
{
if ($v1[$i] != $v2[$i])
return false;
}
}
// If every element matches
return true;
}
// Driver code
$a = array(array( 1, 2, 3 ),
array( 4, 5, 6 ),
array( 7, 8, 9 ));
$b = array(array( 1, 4, 7 ),
array( 2, 5, 6 ),
array( 3, 8, 9 ));
if (check($a, $b))
echo "Yes";
else
echo "No";
// This code is contributed by mits
?>输出：
Yes
时间复杂度： O(N * M * log(max(N，M)))
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