用于查找矩阵中奇数和偶数频率的Java程序

矩阵只是一个二维数组。就像在一维遍历操作中需要弄清楚一样。任务是找到给定矩阵中奇数和偶数的频率。

The size of matrix is defined as product of numbers of columns and rows. So, the size of the matrix is m × n 

这里在矩阵

• M = 矩阵中的行数
• N = 列数

例子：

Input : M = 3
    N = 5
Output: Odd Number Frequency  = 9
    Even Number Frequency = 6
    
Input : M = 3
    N = 3
    {{1, 2, 5},
     {6, 7, 9},
     {3, 2, 1}}
Output: Odd Number Frequency  = 6
    Even Number Frequency = 3    

方法

• 将两个计数器初始化为偶数 = 0 & 奇数 = 0
• 遍历矩阵中存在的每个元素并检查是否 (element % 2 == 0)
• 如果是，则增加 even++
• 否则增加奇数++

下面是该方法在Java中的实现

Java

// Importing Classes/Files
import java.io.*;
  
class GFG {
  
    // Function to compute frequency of odd/even numbers
    public static void findFreq(int[][] arr, int m, int n)
    {
        // Initializing the counter variables
        int even = 0, odd = 0;
  
        // Nested if loops
        // Traversing through each
        // element in the matrix
        for (int i = 0; i < m; i++) {
            
            for (int j = 0; j < n; j++) {
  
                // Checking if the element
                // is divisible by 2
                if (arr[i][j] % 2 == 0) {
  
                    // Increment even counter
                    even++;
                }
                else {
  
                    // Increment odd counter
                    odd++;
                }
            }
        }
  
        // Printing Counts of Enen 
        // and odd numbers in matrix
        System.out.println("Odd Number Frequency: " + odd);
        System.out.println("Even Number Frequence: "
                           + even);
    }
  
    // Main Driver Method
    public static void main(String[] args)
    {
        // Providing inputs to the matrix
        int m = 3, n = 5;
        // Here, m = Number of rows,
        // n = Number  of columns
  
        // Entering elements in the matrix
        int[][] arr = { { 3, 4, 5, 6, 3 },
                        { 4, 3, 2, 7, 9 },
                        { 1, 5, 7, 2, 4 } };
  
        // Calling function to count frequency by
        // passing inputs to the function findFreq
        findFreq(arr, m, n);
    }
}

输出

Odd Number Frequency: 9
Even Number Frequence: 6

时间复杂度：当遍历一维数组时，它取决于执行操作的数组的大小，因为其余执行要么具有相同的顺序，要么在内存（变量）中占用一些恒定空间。这里遍历行的所有元素以及矩阵中的每一行。所以它取决于行和列的乘积。因此，时间复杂度为阶（度）

空间复杂度：只是将内存分配给变量，然后在变量的范围内释放。因为没有为要执行的操作创建其他辅助空间。因此，所需的空间是恒定的，然后释放。默认分配 1，因此空间复杂度为 1。

Time Complexity  = O(Rows*Columns) or O(m * n)
Space Complexity = O(1)