Question

我正在尝试完成一个问题，我将文件读入程序并输出一个文件，其中包含平均值，最小值，最大值以及该数字在程序中出现次数的计数。但是，我无法弄清楚如何为重复数量的＆＃34;计数＆＃34;创建一个数组。

如果我尝试读取的文件的值为19 5 26 5 5 19 16 8 1，

我需要输出的文件来阅读5---3 times; 8---1 time; 16---1 time; 19--2 times; 26--1 times。

我首先将我的数组排序为5 5 5 8 16 19 19 26。

下面是我的代码，解释了我想要做的事情：

#include <iostream>
#include <fstream>
#include <cstdlib>
#include <string>
using namespace std;

double averageArray(int a[], int length); // user defined function to get average
int maxval(int a[], int length); //user defined function to get max val
int minval(int a[], int length); //user defined function to get min val
void bsort(int arr[], int length);// udf to sort array from min to max
int countArray(int a[], int length); //attempt to create a function to get the number of occurrences of a number that is duplicated

int main()
{
    char infilename[16];
    int nums[50];//newly created array to read in the numbers from the file
    int length(0);//array has a length defined by "length"
    ifstream fin;
    ofstream fout;


    cout << "Please enter an input file name: ";
    cin >> infilename;
    cout << endl;

    fin.open(infilename); 
    if (fin.fail())
    {
        cerr << "The file " << infilename << " can't be open!"<<endl;
        return 1;
    }


    cout<<"The output to the file statistics.txt should be as follows: "<<endl;
    fout.open("statistics.txt");
    fout<<"N"<<"\t"<<"Count"<<endl;
    cout<<"N"<<"\t"<<"Count"<<endl;


    while (fin >> nums[length]) 
        length++;

    bsort(nums, length);
    for (int i=0; i<length; i++) {
        if (nums[i]==nums[i-1]) {
            continue;
        }
        cout<<nums[i]<<"\t"<<countArray(nums,length)<<endl;
        fout<<nums[i]<<"\t"<<endl;
    }

    cout << "\nAverage: " << averageArray(nums,length) << endl;
    cout << "Max: "<< maxval(nums,length)<<endl;
    cout << "Min: "<< minval(nums,length)<<endl;


    fin.close();


    return 0;
}



double averageArray (int a[], int length)
{
    double result(0);

    for (int i = 0; i < length ; i++)
        result += a[i];
    return result/length;
}

int maxval(int a[], int length) 
{

    int max(0);

    for (int i=1; i<length; i++)
    {
        if (a[i]>max)
            max=a[i];
    }
    return max;
}

int minval(int a[], int length) 
{

    int min(100);

    for (int i=1; i<length; i++)
    {
        if (a[i]<min)
            min=a[i];
    }
    return min;
}

void bsort(int a[], int length)
{
    for (int i=length-1; i>0; i--)
        for (int j=0; j<i; j++)
            if (a[j]>a[j+1])
            {
                int temp=a[j+1];
                a[j+1]=a[j];
                a[j]=temp;
            }
}

int countArray(int a[], int length)
{
    int counter(0);
    for (int i=0; i<length; i++){
        if (a[i]==a[i+1]) //loop through array and if the number after is the same as the previous number, then count one
        counter++;
    }
    return counter;
}

虽然它编译，但计数仅显示＆＃34; 3＆＃34; s，如下图所示：

output image 。

Answer 1

在我给你解决方案之前，请花一点时间记住，你是用C ++编程，而不是C.因此，你应该使用向量，istream迭代器和std::sort。您还应该使用std::map，这很容易实现此目的：

template <typename It>
std::map<int, int> count_occurrences(It it, It end)
{
  std::map<int, int> output;
  while (it != end) output[*it++]++;
  return output;
}

如何将其与现有代码相结合，留给读者练习。我建议你应该读一下迭代器。

Answer 2

您的函数int countArray(int a[], int length)没有输入实际数字。它始终计算阵列中相同数字的频率。这种情况在五次发生两次，一次发生在19 => 3次。

解决方案：

int countArray(int a[], int length, int num)
{
    int counter(0);
    for (int i=0; i<length; i++){
        if (a[i]==num) //loop through array and if the number is the one you are looking for
           counter++;
    }
    return counter;
}

并致电您：countArray(nums, length, nums[i]);

Answer 3

void countArray(int a[], int length)
{
    int counter(1);
    bool flag(false);
    //you take (i+1) index, it can go out of range
    for (int i = 0; i < length - 1; i++){
        if (a[i]==a[i+1]){ //loop through array and if the number after is the same as the previous number, then count one
            flag = true;
            counter++;
        }
        else {
            if (flag){
                cout<<a[i] << counter << endl;
            }
            flag = false;
            counter = 1;
        }
    }
}

我很久没有在C上编码，但我希望它会有所帮助。此程序将为您打印答案，您只需拨打一次。

Answer 4

我建议使用std::map这是解决问题的最佳解决方案。我将尝试轻松解释执行此操作的不同步骤：

我认为您的变量已初始化，例如：

int length = 9;
int nums[length] = {19, 5, 26, 5, 5, 19, 16, 8, 1};

创建std::map<int,int>，其中key（第一个 int）将是您的号码，value（秒 int）密钥中此数字存储的出现次数。
```
std::map<int,int> listNumber;
```

填写地图

// For all numbers read in your file
for(int i=0; i<length; ++i)
{
    // Get number value
    int n = nums[i];

    // Find number in map
    std::map<int, int>::iterator it = listNumber.find(n);

    // Doesn't exists in map, add it with number of occurence set to 1...
    if(it == listNumber.end())
    {
        listNumber.insert(std::pair<int,int>(n,1));
    }
    // ... otherwise add one to the number of occurence of this number
    else
    {
        it->second = it->second+1;
    }
}

阅读地图

// Read all numbers and display the number of occurence
std::cout << "N" << "\t" << "Count" << std::endl;
for(std::map<int, int>::iterator it = listNumber.begin(); it!=listNumber.end(); ++it)
{
    std::cout << it->first << "\t" << it->second << std::endl;
}

如何计算排序数组中的重复数字

4 个答案: