Find the largest K numbers from array （找出数组中最大的K个值）

Recently i was doing some study on algorithms. A classic problem is to find the K largest(smallest) numbers from an array. I mainly studyed two methods, one is directly methold. It is an extension of select sort, always select the largest number from the array. The pseudo code is as below. The algorithm complexity is O(kn).

 function select(list[1..n], k)
     for i from 1 to k
         minIndex = i
         minValue = list[i]
         for j from i+1 to n
             if list[j] < minValue
                 minIndex = j
                 minValue = list[j]
         swap list[i] and list[minIndex]
     return list[k]

The C++ implementation is

template<typename T>
std::vector<T> SelectLargestKItem(const std::vector<T> &vecInput, size_t K, std::vector<int> &vecIndex)
{
    if (K > vecInput.size())
        return vecInput;

    std::vector<T> vecLocal(vecInput);
    std::vector<T> vecResult;
    for (size_t k = ; k < K; ++ k)
    {
        T maxValue = vecLocal[k];
        int maxIndex = k;
        for (size_t i = k + ; i < vecLocal.size(); ++i) {
            if (vecLocal[i] > maxValue) {
                maxValue = vecLocal[i];
                maxIndex = i;
            }
        }
        if (maxIndex != k)
            std::swap(vecLocal[maxIndex], vecLocal[k]);
        vecResult.push_back( maxValue );
        vecIndex.push_back( maxIndex );
    }
    return vecResult;
}

When the total number of N is very large, such as N > 200,000. And the numbers need to select K is larger than 20, then the above algorithm will become time consuming. After do some research, i choose another algorithm to do the job. This method is a extension of heap sort. The steps work as below:

1) Build a Min Heap MH of the first k elements (arr[0] to arr[k-1]) of the given array. O(k)

2) For each element, after the kth element (arr[k] to arr[n-1]), compare it with root of MH.
……a) If the element is greater than the root then make it root and call heapifyfor MH
……b) Else ignore it.
// The step 2 is O((n-k)*logk)

3) Finally, MH has k largest elements and root of the MH is the kth largest element.

Time Complexity: O(k + (n-k)Logk) without sorted output. If sorted output is needed then O(k + (n-k)Logk + kLogk).

The C++ implementation of the method is as below:

// To heapify a subtree rooted with node i which is
// an index in arr[]. n is size of heap
template<typename T>
void heapifyMinToRoot(std::vector<T> &vecInput, const int n, const int i, std::vector<int> &vecIndex)
{
    int smallestIndex = i;  // Initialize largest as root
    int l =  * i + ;  // left = 2*i + 1
    int r =  * i + ;  // right = 2*i + 2

    // If left child is larger than root
    if (l < n && vecInput[l] < vecInput[smallestIndex])
        smallestIndex = l;

    // If right child is larger than largest so far
    if (r < n && vecInput[r] < vecInput[smallestIndex])
        smallestIndex = r;

    // If largest is not root
    if (smallestIndex != i)
    {
        std::swap(vecInput[i], vecInput[smallestIndex]);
        std::swap(vecIndex[i], vecIndex[smallestIndex]);

        // Recursively heapify the affected sub-tree
        heapifyMinToRoot(vecInput, n, smallestIndex, vecIndex);
    }
}

template<typename T>
std::vector<T> SelectLargestKItemHeap(const std::vector<T> &vecInput, const size_t K, std::vector<int> &vecIndex)
{
    if (K > vecInput.size())  {
        std::vector<T> vecResult(vecInput);
        std::sort(vecResult.begin(), vecResult.end());
        std::reverse(vecResult.begin(), vecResult.end());
        for (size_t i = ; i < vecInput.size(); ++i)
            vecIndex.push_back(i);
        return vecResult;
    }

    std::vector<T> vecLocal(vecInput);
    std::vector<T> vecResult(vecInput.begin(), vecInput.begin() + K);
    vecIndex.clear();
    for (size_t i = ; i < K; ++ i) vecIndex.push_back(i);

    for (int K1 = K /  - ; K1 >= ; -- K1)
        heapifyMinToRoot(vecResult, K, K1, vecIndex);

    for (size_t i = K; i < vecLocal.size(); ++ i) {
        if (vecLocal[i] > vecResult[]) {
            vecResult[] = vecLocal[i];
            vecIndex[] = i;

            for (int K1 = K /  - ; K1 >= ; -- K1)
                heapifyMinToRoot(vecResult, K, K1, vecIndex);
        }
    }
    for (int k = K - ; k >= ; -- k )
    {
        std::swap(vecResult[k], vecResult[]);
        std::swap(vecIndex[k], vecIndex[]);

        heapifyMinToRoot(vecResult, k, , vecIndex);
    }

    return vecResult;
}

Here is the code to test these two methods.

void SelectionAlgorithmBenchMark()
{
    int N = ;
    std::vector<int> vecInput;

    std::minstd_rand0 generator();
    for (int i = ; i < N; ++i)
    {
        int nValue = generator();
        vecInput.push_back(nValue );
    }
    std::vector<int> vecResult, vecIndex;
    int K = ;
    CStopWatch stopWatch;
    vecResult = SelectLargestKItem<int>(vecInput, K, vecIndex);
    std::cout << "Standard algorithm SelectLargestKItem takes " << stopWatch.Now() << " ms" << std::endl;
    for (int k = ; k < K; ++k)
    {
        std::cout << "Index " << vecIndex[k] << ", value " << vecResult[k] << std::endl;
    }
    std::cout << std::endl;

    stopWatch.Start();
    vecResult = SelectLargestKItemHeap<int>(vecInput, K, vecIndex);
    std::cout << "Heap algorithm SelectLargestKItem takes " << stopWatch.Now() << " ms" << std::endl;
    for (int k = ; k < K; ++k)
    {
        std::cout << "Index " << vecIndex[k] << ", value " << vecResult[k] << std::endl;
    }
}

When N is 200000, K is 20, the first method takes 353ms, the second method takes 31ms. The difference is more than 10 times.