在给定的矩阵中查找山脉的数量

鉴于大小N×N的二维方阵，任务是统计在矩阵山脉的数量。

An element in a matrix is said to be a mountain when all the surrounding elements (in all 8 directions) are smaller than the given element.

例子：

Input: matrix = 
{ { 4, 5, 6 },
  { 2, 1, 3 },
  { 7, 8, 9 } }
Output: 1
Explanation 
Only mountain element = 9. 
All the neighbouring elements 
1, 3 and 8 are smaller than 9.

Input: matrix = 
{ { 7, 8, 9 },
  { 1, 2, 3 },
  { 4, 5, 6 } }
Output: 2
Explanation
Mountain elements = 6 (2, 3 and 5)
and 9 (8, 2, and 3) 

方法：想法是遍历矩阵，同时检查所有可能的 8 个方向上的相邻元素。如果元素大于所有元素，则增加计数器变量。最后，返回计数器。

1. 创建一个大小为(N + 2) X (N + 2)的辅助数组。
2. 用INT_MIN值填充所有边框元素
3. 在NXN剩余的数组空间中，复制原矩阵
4. 现在检查一个元素是否大于所有 8 个方向的元素。
5. 计算这些元素的数量并打印出来。

例如：

If matrix = 
{ { 7, 8, 9 },
  { 1, 2, 3 },
  { 4, 5, 6 } }

The auxiliary array would be
{ { 0, 0, 0, 0, 0 },
  { 0, 7, 8, 9, 0 },
  { 0, 1, 2, 3, 0 },
  { 0, 4, 5, 6, 0 },
  { 0, 0, 0, 0, 0 } }

以下是上述方法的实现：

C++

// C++ program find the count of
// mountains in a given Matrix
 
#include 
using namespace std;
 
const int MAX = 100;
 
// Function to count number of mountains
// in a given matrix of size n
int countMountains(int a[][MAX], int n)
{
    int A[n + 2][n + 2];
    int count = 0;
 
    // form another matrix with one extra
    // layer of border elements. Border
    // elements will contain INT_MIN value.
    for (int i = 0; i < n + 2; i++) {
        for (int j = 0; j < n + 2; j++) {
 
            if ((i == 0) || (j == 0)
                || (i == n + 1)
                || (j == n + 1)) {
 
                // For border elements,
                // set value as INT_MIN
                A[i][j] = INT_MIN;
            }
            else {
 
                // For rest elements, just copy
                // it into new matrix
                A[i][j] = a[i - 1][j - 1];
            }
        }
    }
 
    // Check for mountains in the modified matrix
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
 
            // check for all directions
            if ((A[i][j] > A[i - 1][j])
                && (A[i][j] > A[i + 1][j])
                && (A[i][j] > A[i][j - 1])
                && (A[i][j] > A[i][j + 1])
                && (A[i][j] > A[i - 1][j - 1])
                && (A[i][j] > A[i + 1][j + 1])
                && (A[i][j] > A[i - 1][j + 1])
                && (A[i][j] > A[i + 1][j - 1])) {
                count++;
            }
        }
    }
 
    return count;
}
 
// Driver code
int main()
{
    int a[][MAX] = { { 1, 2, 3 },
                     { 4, 5, 6 },
                     { 7, 8, 9 } };
    int n = 3;
 
    cout << countMountains(a, n);
    return 0;
}

Java

// Java program find the count of
// mountains in a given Matrix
import java.util.*;
 
class GFG{
  
static int MAX = 100;
  
// Function to count number of mountains
// in a given matrix of size n
static int countMountains(int a[][], int n)
{
    int [][]A = new int[n + 2][n + 2];
    int count = 0;
  
    // form another matrix with one extra
    // layer of border elements. Border
    // elements will contain Integer.MIN_VALUE value.
    for (int i = 0; i < n + 2; i++) {
        for (int j = 0; j < n + 2; j++) {
  
            if ((i == 0) || (j == 0)
                || (i == n + 1)
                || (j == n + 1)) {
  
                // For border elements,
                // set value as Integer.MIN_VALUE
                A[i][j] = Integer.MIN_VALUE;
            }
            else {
  
                // For rest elements, just copy
                // it into new matrix
                A[i][j] = a[i - 1][j - 1];
            }
        }
    }
  
    // Check for mountains in the modified matrix
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
  
            // check for all directions
            if ((A[i][j] > A[i - 1][j])
                && (A[i][j] > A[i + 1][j])
                && (A[i][j] > A[i][j - 1])
                && (A[i][j] > A[i][j + 1])
                && (A[i][j] > A[i - 1][j - 1])
                && (A[i][j] > A[i + 1][j + 1])
                && (A[i][j] > A[i - 1][j + 1])
                && (A[i][j] > A[i + 1][j - 1])) {
                count++;
            }
        }
    }
  
    return count;
}
  
// Driver code
public static void main(String[] args)
{
    int a[][] = { { 1, 2, 3 },
                     { 4, 5, 6 },
                     { 7, 8, 9 } };
    int n = 3;
  
    System.out.print(countMountains(a, n));
}
}
 
// This code is contributed by sapnasingh4991

Python3

# Python3 program find the count of
# mountains in a given Matrix
MAX = 100
 
# Function to count number of mountains
# in a given matrix of size n
def countMountains(a, n):
    A = [[0 for i in range(n+2)] for i in range(n+2)]
    count = 0
 
    # form another matrix with one extra
    # layer of border elements. Border
    # elements will contain INT_MIN value.
    for i in range(n+2):
        for j in range(n+2):
            if ((i == 0) or (j == 0) or
                (i == n + 1) or (j == n + 1)):
 
                # For border elements,
                # set value as INT_MIN
                A[i][j] = float('-inf')
            else:
 
                # For rest elements, just copy
                # it into new matrix
                A[i][j] = a[i - 1][j - 1]
             
    # Check for mountains in the modified matrix
    for i in range(n + 1):
        for j in range(n + 1):
            if ((A[i][j] > A[i - 1][j]) and
                (A[i][j] > A[i + 1][j]) and
                (A[i][j] > A[i][j - 1]) and
                (A[i][j] > A[i][j + 1]) and
                (A[i][j] > A[i - 1][j - 1]) and
                (A[i][j] > A[i + 1][j + 1]) and
                (A[i][j] > A[i - 1][j + 1]) and
                (A[i][j] > A[i + 1][j - 1])):
                count = count + 1
 
    return count
 
# Driver code
a = [ [ 1, 2, 3 ],
    [ 4, 5, 6 ],
    [ 7, 8, 9 ] ]
 
n = 3
 
print(countMountains(a, n))
 
# This code is contributed by Sanjit_Prasad

C#

// C# program find the count of
// mountains in a given Matrix
using System;
 
class GFG{
 
    static int MAX = 100;
     
    // Function to count number of mountains
    // in a given matrix of size n
    static int countMountains(int [,]a, int n)
    {
        int [,]A = new int[n + 2,n + 2];
        int count = 0;
     
        // form another matrix with one extra
        // layer of border elements. Border
        // elements will contain Integer.MIN_VALUE value.
        for (int i = 0; i < n + 2; i++) {
            for (int j = 0; j < n + 2; j++) {
     
                if ((i == 0) || (j == 0)
                    || (i == n + 1)
                    || (j == n + 1)) {
     
                    // For border elements,
                    // set value as Integer.MIN_VALUE
                    A[i,j] = int.MinValue;
                }
                else {
     
                    // For rest elements, just copy
                    // it into new matrix
                    A[i,j] = a[i - 1,j - 1];
                }
            }
        }
     
        // Check for mountains in the modified matrix
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= n; j++) {
     
                // check for all directions
                if ((A[i,j] > A[i - 1,j])
                    && (A[i,j] > A[i + 1,j])
                    && (A[i,j] > A[i,j - 1])
                    && (A[i,j] > A[i,j + 1])
                    && (A[i,j] > A[i - 1,j - 1])
                    && (A[i,j] > A[i + 1,j + 1])
                    && (A[i,j] > A[i - 1,j + 1])
                    && (A[i,j] > A[i + 1,j - 1])) {
                    count++;
                }
            }
        }
     
        return count;
    }
     
    // Driver code
    public static void Main(string[] args)
    {
        int [,]a = { { 1, 2, 3 },
                        { 4, 5, 6 },
                        { 7, 8, 9 } };
        int n = 3;
     
        Console.WriteLine(countMountains(a, n));
    }
}
 
// This code is contributed by AnkitRai01

Javascript

//Javascript program find the count of
// mountains in a given Matrix
 
// Function to count number of mountains
// in a given matrix of size n
function countMountains(a,  n)
{
    var A= new Array(n+2).fill(0).map(() => new Array(n+2).fill(0));
    var count = 0;
 
    // form another matrix with one extra
    // layer of border elements. Border
    // elements will contain INT_MIN value.
    for (var i = 0; i < n + 2; i++) {
        for (var j = 0; j < n + 2; j++) {
 
            if ((i == 0) || (j == 0)
                || (i == n + 1)
                || (j == n + 1)) {
 
                // For border elements,
                // set value as INT_MIN
                A[i][j] = Number.MIN_VALUE;
            }
            else {
 
                // For rest elements, just copy
                // it into new matrix
                A[i][j] = a[i - 1][j - 1];
            }
        }
    }
 
    // Check for mountains in the modified matrix
    for (var i = 1; i <= n; i++) {
        for (var j = 1; j <= n; j++) {
 
            // check for all directions
            if ((A[i][j] > A[i - 1][j])
                && (A[i][j] > A[i + 1][j])
                && (A[i][j] > A[i][j - 1])
                && (A[i][j] > A[i][j + 1])
                && (A[i][j] > A[i - 1][j - 1])
                && (A[i][j] > A[i + 1][j + 1])
                && (A[i][j] > A[i - 1][j + 1])
                && (A[i][j] > A[i + 1][j - 1])) {
                count++;
            }
        }
    }
 
    return count;
}
 
var a = [[ 1, 2, 3 ],
                     [ 4, 5, 6 ],
                     [ 7, 8, 9 ] ];
var n = 3;
  document.write( countMountains(a, n));
 
 
 
//This code is contributed by SoumikMondal

输出：

1

性能分析：

• 时间复杂度：在上述方法中，我们进行了两次迭代。第一个是在(N + 2) X (N + 2) 个元素上创建辅助矩阵。第二个在NXN元素上找到实际的山元素，所以时间复杂度是O(NXN) 。
• 辅助空间复杂度：在上述方法中，我们使用大小为(N + 2) X (N + 2)的辅助矩阵，因此辅助空间复杂度为O(N *N) 。

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