计算矩阵中所有已排序的行
给定一个 m*n 大小的矩阵，任务是计算矩阵中按严格升序或严格降序排序的所有行？
例子 ：
Input : m = 4, n = 5
mat[m][n] = 1 2 3 4 5
4 3 1 2 6
8 7 6 5 4
5 7 8 9 10
Output: 3 这个想法很简单，涉及矩阵的两次遍历。
1) 从矩阵的左边开始遍历，统计所有严格递增的行
2）从矩阵的右侧开始遍历，计算所有严格降序的行
下面是上述想法的实现。
C++
// C++ program to find number of sorted rows
#include
#define MAX 100
using namespace std;
// Function to count all sorted rows in a matrix
int sortedCount(int mat[][MAX], int r, int c)
{
int result = 0; // Initialize result
// Start from left side of matrix to
// count increasing order rows
for (int i=0; i0; j--)
if (mat[i][j-1] <= mat[i][j])
break;
// Note c > 1 condition is required to make
// sure that a single column row is not counted
// twice (Note that a single column row is sorted
// both in increasing and decreasing order)
if (c > 1 && j == 0)
result++;
}
return result;
}
// Driver program to run the case
int main()
{
int m = 4, n = 5;
int mat[][MAX] = {{1, 2, 3, 4, 5},
{4, 3, 1, 2, 6},
{8, 7, 6, 5, 4},
{5, 7, 8, 9, 10}};
cout << sortedCount(mat, m, n);
return 0;
} Java
// Java program to find number of sorted rows
class GFG {
static int MAX = 100;
// Function to count all sorted rows in a matrix
static int sortedCount(int mat[][], int r, int c)
{
int result = 0; // Initialize result
// Start from left side of matrix to
// count increasing order rows
for (int i = 0; i < r; i++) {
// Check if there is any pair ofs element
// that are not in increasing order.
int j;
for (j = 0; j < c - 1; j++)
if (mat[i][j + 1] <= mat[i][j])
break;
// If the loop didn't break (All elements
// of current row were in increasing order)
if (j == c - 1)
result++;
}
// Start from right side of matrix to
// count increasing order rows ( reference
// to left these are in decreasing order )
for (int i = 0; i < r; i++) {
// Check if there is any pair ofs element
// that are not in decreasing order.
int j;
for (j = c - 1; j > 0; j--)
if (mat[i][j - 1] <= mat[i][j])
break;
// Note c > 1 condition is required to make
// sure that a single column row is not counted
// twice (Note that a single column row is sorted
// both in increasing and decreasing order)
if (c > 1 && j == 0)
result++;
}
return result;
}
// Driver code
public static void main(String arg[])
{
int m = 4, n = 5;
int mat[][] = { { 1, 2, 3, 4, 5 },
{ 4, 3, 1, 2, 6 },
{ 8, 7, 6, 5, 4 },
{ 5, 7, 8, 9, 10 } };
System.out.print(sortedCount(mat, m, n));
}
}
// This code is contributed by Anant Agarwal.Python
# Python3 program to find number
# of sorted rows
def sortedCount(mat, r, c):
result = 0
# Start from left side of matrix to
# count increasing order rows
for i in range(r):
# Check if there is any pair ofs element
# that are not in increasing order.
j = 0
for j in range(c - 1):
if mat[i][j + 1] <= mat[i][j]:
break
# If the loop didn't break (All elements
# of current row were in increasing order)
if j == c - 2:
result += 1
# Start from right side of matrix to
# count increasing order rows ( reference
# to left these are in decreasing order )
for i in range(0, r):
# Check if there is any pair ofs element
# that are not in decreasing order.
j = 0
for j in range(c - 1, 0, -1):
if mat[i][j - 1] <= mat[i][j]:
break
# Note c > 1 condition is required to
# make sure that a single column row
# is not counted twice (Note that a
# single column row is sorted both
# in increasing and decreasing order)
if c > 1 and j == 1:
result += 1
return result
# Driver code
m, n = 4, 5
mat = [[1, 2, 3, 4, 5],
[4, 3, 1, 2, 6],
[8, 7, 6, 5, 4],
[5, 7, 8, 9, 10]]
print(sortedCount(mat, m, n))
# This code is contributed by
# Mohit kumar 29 (IIIT gwalior)C#
// C# program to find number of sorted rows
using System;
class GFG {
// static int MAX = 100;
// Function to count all sorted rows in
// a matrix
static int sortedCount(int [,]mat, int r,
int c)
{
int result = 0; // Initialize result
// Start from left side of matrix to
// count increasing order rows
for (int i = 0; i < r; i++) {
// Check if there is any pair of
// element that are not in
// increasing order.
int j;
for (j = 0; j < c - 1; j++)
if (mat[i,j + 1] <= mat[i,j])
break;
// If the loop didn't break (All
// elements of current row were
// in increasing order)
if (j == c - 1)
result++;
}
// Start from right side of matrix
// to count increasing order rows
// ( reference to left these are in
// decreasing order )
for (int i = 0; i < r; i++) {
// Check if there is any pair
// ofs element that are not in
// decreasing order.
int j;
for (j = c - 1; j > 0; j--)
if (mat[i,j - 1] <= mat[i,j])
break;
// Note c > 1 condition is
// required to make sure that a
// single column row is not
// counted twice (Note that a
// single column row is sorted
// both in increasing and
// decreasing order)
if (c > 1 && j == 0)
result++;
}
return result;
}
// Driver code
public static void Main()
{
int m = 4, n = 5;
int [,]mat = { { 1, 2, 3, 4, 5 },
{ 4, 3, 1, 2, 6 },
{ 8, 7, 6, 5, 4 },
{ 5, 7, 8, 9, 10 } };
Console.WriteLine(
sortedCount(mat, m, n));
}
}
// This code is contributed by anuj_67.PHP
0; $j--)
if ($mat[$i][$j - 1] <= $mat[$i][$j])
break;
// Note c > 1 condition is
// required to make sure that
// a single column row is not
// counted twice (Note that a
// single column row is sorted
// both in increasing and
// decreasing order)
if ($c > 1 && $j == 0)
$result++;
}
return $result;
}
// Driver Code
$m = 4; $n = 5;
$mat = array(array(1, 2, 3, 4, 5),
array(4, 3, 1, 2, 6),
array(8, 7, 6, 5, 4),
array(5, 7, 8, 9, 10));
echo sortedCount($mat, $m, $n);
// This code is contributed by anuj_67.
?>Javascript
输出 ：
3