c++ - 查找数组中的出现

Question

我必须找到数组中每个元素的出现。到目前为止我的代码是这样的

void Occurrences()
{
    int numers[10], count = 0, i;
    for (i = 0; i < 10; i++)
    {
        cout << "Enter Number";
        cin >> numers[i];
    }
    for (i = 0; i < 10; i++)
    {
        for (int j = 0; j < 10; j++)
        {
            if (numers[i] == numers[j])
            {
                count++;
            }
        }
        cout << numers[i] << " is Occur " << count << " Time in Array" << endl;
        count = 0;
    }
}
int main()
{
    Occurrences();
}

输出将乘以相同的数字，即如果我输入六个 1 和 4 个 2。输出是

1 is occur 6 time in array.
1 is occur 6 time in array.
1 is occur 6 time in array.
1 is occur 6 time in array.
1 is occur 6 time in array.
1 is occur 6 time in array.
2 is occur 4 time in array.
2 is occur 4 time in array.
2 is occur 4 time in array.
2 is occur 4 time in array.

但我想要这样的输出:

1 is occur 6 time in array.
2 is occur 4 time in array.

我该怎么做？

Answer 1

既然你标记了这个 C++11，我会使用 std::unordered_map :

void Occurrences()
{
    std::unordered_map<int, int> occurrences;

    // enter 10 numbers
    for (int i = 0; i < 10; ++i) {
        cout << "Enter Number";
        int x;
        cin >> x;
        ++occurrences[x]; // increment the count for x
    }

    // print results
    for (const auto& pr : occurrences) {
        std::cout << pr.first << " appears " << pr.second << " times." << std::endl;
    }
}