矩阵中所有奇数频率元素的总和

给定包含重复元素的 NxM 整数矩阵。任务是找到给定矩阵中所有奇数出现元素的总和。这是矩阵中频率为奇数的所有此类元素的总和。
例子：

Input : mat[] = {{1, 1, 2},
                {2, 3, 3},
                {4, 5, 3}}
Output : 18
The odd occurring elements are 3, 4, 5 and their number
of occurrences are 3, 1, 1 respectively. Therefore,
sum = 3+3+3+4+5 = 18.

Input : mat[] = {{10, 20},
                 {40, 40}}
Output : 30

方法：

• 遍历矩阵并使用C++中的map来存储矩阵元素的频率，使得map的key是矩阵元素，value是它在矩阵中的频率。
• 然后，遍历地图查找元素的频率并检查它是否是奇数，如果是奇数，则将这个元素的频率乘以求和。

下面是上述方法的实现：

C++

// C++ program to find sum of all odd
// frequency elements in a Matrix
 
#include 
using namespace std;
 
#define N 3 // Rows
#define M 3 // Columns
 
// Function to find sum of all odd
// frequency elements in a Matrix
int sumOddOccurring(int arr[N][M])
{
 
    // Store frequencies of elements
    // in matrix
    map mp;
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < M; j++) {
            mp[arr[i][j]]++;
        }
    }
 
    // Sum of odd frequency elements
    int sum = 0;
    for (auto itr = mp.begin(); itr != mp.end(); itr++) {
        if (itr->second % 2 != 0) {
            sum += (itr->first) * (itr->second);
        }
    }
 
    return sum;
}
 
// Driver Code
int main()
{
 
    int mat[N][M] = { { 1, 2, 3 },
                      { 1, 3, 2 },
                      { 1, 5, 6 } };
 
    cout << sumOddOccurring(mat) << endl;
 
    return 0;
}

Java

// Java program to find sum of all odd
// frequency elements in a Matrix
 
import java.util.*;
 
class GFG
{
 
    static int N = 3; // Rows
    static int M = 3; // Columns
 
    // Function to find sum of all odd
    // frequency elements in a Matrix
    static int sumOddOccurring(int arr[][])
    {
        // Store frequencies of elements
        // in matrix
        Map mp = new HashMap<>();
        for (int i = 0; i < N; i++)
        {
            for (int j = 0; j < M; j++)
            {
                if (mp.containsKey(arr[i][j]))
                {
                    mp.put(arr[i][j], mp.get(arr[i][j]) + 1);
                }
                else
                {
                    mp.put(arr[i][j], 1);
                }
            }
        }
 
        int sum = 0;
         
        // Sum of odd frequency elements
        for (Map.Entry itr : mp.entrySet())
        {
            if (itr.getValue() % 2 != 0)
            {
                sum += (itr.getKey()) * (itr.getValue());
            }
        }
 
        return sum;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int mat[][] = {{1, 2, 3},
        {1, 3, 2},
        {1, 5, 6}};
 
        System.out.println(sumOddOccurring(mat));
    }
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 program to find sum of all odd
# frequency elements in a Matrix
 
# Function to find sum of all odd
# frequency elements in a Matrix
def sumOddOccurring(mat):
     
    # Store frequencies of elements
    # in matrix
    mp = {}
    n, m = len(mat), len(mat[0])
    for i in range(n):
        for j in range(m):
            if mat[i][j] in mp:
                mp[mat[i][j]] = mp.get(mat[i][j]) + 1
            else:
                mp[mat[i][j]] = 1
 
    # Sum of odd frequency elements
    _sum = 0
    for i in range(n):
        for j in range(m):
            if mp.get(mat[i][j]) % 2 == 1:
                _sum+=mat[i][j]
    return _sum
 
# Driver Code
if __name__=='__main__':
    mat=[[1,2,3],[1,3,2],[1,5,6]]
    print(sumOddOccurring(mat))
 
# This code is Contributed by Vikash Kumar 37

C#

// C# program to find sum of all odd
// frequency elements in a Matrix
using System;
using System.Collections.Generic;
 
class GFG
{
 
    static int N = 3; // Rows
    static int M = 3; // Columns
 
    // Function to find sum of all odd
    // frequency elements in a Matrix
    static int sumOddOccurring(int [,]arr)
    {
        // Store frequencies of elements
        // in matrix
        Dictionary mp = new Dictionary();
        for (int i = 0; i < N; i++)
        {
            for (int j = 0; j < M; j++)
            {
                if (mp.ContainsKey(arr[i,j]))
                {
                    var v = mp[arr[i,j]];
                    mp.Remove(arr[i,j]);
                    mp.Add(arr[i,j], ++v);
                }
                else
                {
                    mp.Add(arr[i,j], 1);
                }
            }
        }
 
        int sum = 0;
         
        // Sum of odd frequency elements
        foreach(KeyValuePair itr in mp)
        {
            if (itr.Value % 2 != 0)
            {
                sum += (itr.Key) * (itr.Value);
            }
        }
 
        return sum;
    }
 
    // Driver Code
    public static void Main(String[] args)
    {
        int [,]mat = {{1, 2, 3},
        {1, 3, 2},
        {1, 5, 6}};
 
        Console.WriteLine(sumOddOccurring(mat));
    }
}
 
// This code is contributed by Princi Singh

Javascript

 
// JavaScript program to find sum of all odd
// frequency elements in a Matrix
var N = 3; // Rows
var M = 3; // Columns
// Function to find sum of all odd
// frequency elements in a Matrix
function sumOddOccurring(arr)
{
    // Store frequencies of elements
    // in matrix
    var mp = new Map();
    for (var i = 0; i < N; i++)
    {
        for (var j = 0; j < M; j++)
        {
            if (mp.has(arr[i][j]))
            {
                var v = mp.get(arr[i][j]);
                mp.delete(arr[i][j]);
                mp.set(arr[i][j], ++v);
            }
            else
            {
                mp.set(arr[i][j], 1);
            }
        }
    }
    var sum = 0;
     
    // Sum of odd frequency elements
    mp.forEach((value, key) => {
        if (value % 2 != 0)
        {
            sum += (key) * (value);
        }
    });
    return sum;
}
// Driver Code
var mat = [[1, 2, 3],
[1, 3, 2],
[1, 5, 6]];
document.write(sumOddOccurring(mat));
 
 

输出：

14

时间复杂度： O(N x M)
辅助空间： O(N x M)

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