在C ++中筛选出数组中的重复值 [英] Filter out duplicate values in array in C++

问题描述

例如，我有一排十个数字:

I have a row of ten numbers for example:

5 5 6 7 5 9 4 2 2 7

5 5 6 7 5 9 4 2 2 7

现在，我想要一个程序，该程序可以查找所有重复项，并在控制台中将它们给出，例如3次5次，2次2次，2次7次.虽然我确实编写了一种算法，该算法可以在一排数字中查找重复项，但是我无法如所述在控制台中将它们给出.我的程序将输出:

Now I want a program that finds all duplicates and gives them out in the console like 3 times 5, 2 times 2, 2 times 7. While I did code an algorithm that finds duplicates in a row of numbers I can't give them out in the console as described. My program will output:

3 times 5
2 times 5
2 times 7
2 times 2

我该如何解决这个问题?

How can I solve this problem?

#include <iostream>

using namespace std;

int main()
{
  int arr[10];
  int i,j;
  int z = 1;

    for(i = 0; i < 10; i++) {
        cin >> arr[i];
    }
    for(i = 0; i < 10; i++){
        for(j = i+1; j < 10; j++){
            if(arr[i] == arr[j]){
                z++;
            }
        }
        if(z >= 2){
        cout << z << " times " << arr[i] << endl;
        z = 1;
        }

    }


    return 0;
}

推荐答案

您需要检查以前是否未找到 arr [i] ，例如:

You need to check that arr[i] is not already found before, like this for example:

if(z >= 2) {
    int found_before = 0;
    for(j = 0; j < i; ++j)
        if(arr[i] == arr[j])
            found_before = 1;
    if(!found_before)
        cout << z << " times " << arr[i] << endl;
    z = 1;
}

将打印:

3 times 5
2 times 7
2 times 2

那样，您就不会再次打印 5 .

That way you don't print 5 again.

使用您的代码，它将打印出它发现5次3次(对于数组中的前5个)，然后当它移至数组中的后5个时，它将忘记数组中的前5个，并报告它两次发现5(自身和数组的第5个数字).

With your code it would print that it found 5 three times (for the first 5 in your array), and then when it would move to he second 5 in your array, it would forgot about the first 5 in your array, and report that it found 5 twice (itself and the 5th number of the array).

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