Question

1       1       5       7       5             
6       3       0       4       0              
6       9       0       4       0             
8       4       3       3       1             
8       2       8       8       0             
7       8       6       4       4              
7       6       4       4       0                   
9       4       2       4       0                  



 void something(vector < vector <int> > v)
{

    sort(v.begin(), v.end());
    int uniqueCount = std::unique(v.begin(), v.end()) - v.begin();
    count = uniqueCount-1;
}

我想计算除0以外的其他数字。在这种情况下，是1 3 4 5 = 4个唯一数字。我没有得到正确的答案，它正在尝试删除矩阵的几行。

Answer 1

我建议使用std::set进行以下实现，该实现将每个值存储一次：

#include <set>
#include <vector>
#include <iostream>

using namespace std;

int count_object(vector<vector<int>> &matrix)
{
    std::set<int> s;
    for(auto &v : matrix)
        s.insert(v.begin(), v.end());
    s.erase(0);
    return s.size();
}

int main()
{
    vector<vector<int>> v = { { 1, 2, 3 }, { 3, 2, 1 }, { 0, 4, 1 } };
    std::cout << "Unique numbers: " << count_object(v);
}

Answer 2

如果由于某种原因您不能使用集合，只需在关于count_if的第一条评论上展开

// flatten the 1D Vector
vector<int> all_ints;
for(auto& currv : N)
{
    all_ints.insert(all_ints.end(), currv.begin(), currv.end());
}

std::map<int,bool> unique_ints_map;
// use a lambda expression to count new elements
int unique_items = std::count_if(all_ints.begin(), all_ints.end(), 
                                [&unique_ints_map](int i)
                                {
                                    // Element exists already so shouldn't be added
                                    if( unique_ints_map.count(i) > 0 )
                                    {
                                        return false;
                                    }
                                    // Throw out zeros too since you don't want to count it
                                    if( i == 0)
                                    {
                                        return false;
                                    }
                                    // This is a new unique number, add it to our map so we don't count again
                                    unique_ints_map[i] = true;
                                    return true;
                                }
                                );
std::cout << "unique_items: " << unique_items << '\n';

如何计算向量中的唯一编号而不删除它？

2 个答案: