C++ 程序 矩阵中偶数和奇数数字的频率

给定一个m×n矩阵，任务是找到矩阵中偶数和奇数数字的频率。

示例:

Input : m = 3, n = 3
        { 1, 2, 3 }, 
        { 4, 5, 6 }, 
        { 7, 8, 9 }
Output : Frequency of odd number =  5 
         Frequency of even number = 4


Input :   m = 3, n = 3
         { 10, 11, 12 },
         { 13, 14, 15 },
         { 16, 17, 18 }
Output : Frequency of odd number  =  4 
         Frequency of even number  = 5

// C++ Program to Find the frequency
// of even and odd numbers in a matrix
# include <bits/stdc++.h>
using namespace std;
 
# define MAX 100
 
// function for calculating frequency
void freq(int ar[][MAX], int m, int n)
{
    int even = 0, odd = 0;
 
    for (int i = 0; i < m; ++i) {
        for (int j = 0; j < n; ++j) {
            // modulo by 2 to check
            // even and odd
            if ((ar[i][j] % 2) == 0)
                ++even;
            else
                ++odd;
        }
    }
 
    // print Frequency of numbers
    cout << "Frequency of odd number = " << odd << endl;
    cout << "Frequency of even number = " << even << endl;
}
 
// Driver code
int main()
{
    int m = 3, n = 3;
 
    int array[][MAX]
        = { { 1, 2, 3 }, { 4, 5, 6 }, { 7, 8, 9 } };
 
    freq(array, m, n);
    return 0;
}  

输出

Frequency of odd number = 5
Frequency of even number = 4

时间复杂度: O(m*n)，其中 m 是给定矩阵中的行数, n 是给定矩阵中的列数。

辅助空间： O(1)

方法：使用递归方法

// C++ Program to Find the frequency
// of even and odd numbers in a matrix
# include <bits/stdc++.h>
using namespace std;
 
# define MAX 100
int even = 0, odd = 0;
 
void freq(int ar[][MAX], int m, int n, int i, int j) {
    if (i == m) //基本条件
    {
        cout << "Frequency of odd number = " << odd << endl;
        cout << "Frequency of even number = " << even << endl;
        return;
    }
    if (j == n)
    {
        freq(ar, m, n, i+1, 0); //递归调用
        return;
    }
    if (ar[i][j] & 1)
        ++odd;
    else
        ++even;
    freq(ar, m, n, i, j+1);  //递归调用
}
 
int main() {
    int m = 3,n = 3;
    int array[][MAX] = { { 1, 2, 3 }, { 4, 5, 6 }, { 7, 8, 9 } };
    freq(array, m, n, 0, 0);
    return 0;
}  

输出

Frequency of odd number = 5
Frequency of even number = 4

时间复杂度: O(m*n)，其中 m 是给定矩阵中的行数, n 是给定矩阵中的列数。

辅助空间： O(m*n) ，每次调用函数freq在调用堆栈中占用O(1)的空间，调用次数为m*n。