首先，我们可以从未排序的数组中构建一个BST，它需要O(n)时间，从BST中我们可以找到O(log(n))中第k个最小的元素，它的总计数为O(n)。

中位数中位数算法的解释可以在这里找到n中第k大的整数: http://cs.indstate.edu/~spitla/presentation.pdf

c++中的实现如下:

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int findMedian(vector<int> vec){
//    Find median of a vector
    int median;
    size_t size = vec.size();
    median = vec[(size/2)];
    return median;
}

int findMedianOfMedians(vector<vector<int> > values){
    vector<int> medians;

    for (int i = 0; i < values.size(); i++) {
        int m = findMedian(values[i]);
        medians.push_back(m);
    }

    return findMedian(medians);
}

void selectionByMedianOfMedians(const vector<int> values, int k){
//    Divide the list into n/5 lists of 5 elements each
    vector<vector<int> > vec2D;

    int count = 0;
    while (count != values.size()) {
        int countRow = 0;
        vector<int> row;

        while ((countRow < 5) && (count < values.size())) {
            row.push_back(values[count]);
            count++;
            countRow++;
        }
        vec2D.push_back(row);
    }

    cout<<endl<<endl<<"Printing 2D vector : "<<endl;
    for (int i = 0; i < vec2D.size(); i++) {
        for (int j = 0; j < vec2D[i].size(); j++) {
            cout<<vec2D[i][j]<<" ";
        }
        cout<<endl;
    }
    cout<<endl;

//    Calculating a new pivot for making splits
    int m = findMedianOfMedians(vec2D);
    cout<<"Median of medians is : "<<m<<endl;

//    Partition the list into unique elements larger than 'm' (call this sublist L1) and
//    those smaller them 'm' (call this sublist L2)
    vector<int> L1, L2;

    for (int i = 0; i < vec2D.size(); i++) {
        for (int j = 0; j < vec2D[i].size(); j++) {
            if (vec2D[i][j] > m) {
                L1.push_back(vec2D[i][j]);
            }else if (vec2D[i][j] < m){
                L2.push_back(vec2D[i][j]);
            }
        }
    }

//    Checking the splits as per the new pivot 'm'
    cout<<endl<<"Printing L1 : "<<endl;
    for (int i = 0; i < L1.size(); i++) {
        cout<<L1[i]<<" ";
    }

    cout<<endl<<endl<<"Printing L2 : "<<endl;
    for (int i = 0; i < L2.size(); i++) {
        cout<<L2[i]<<" ";
    }

//    Recursive calls
    if ((k - 1) == L1.size()) {
        cout<<endl<<endl<<"Answer :"<<m;
    }else if (k <= L1.size()) {
        return selectionByMedianOfMedians(L1, k);
    }else if (k > (L1.size() + 1)){
        return selectionByMedianOfMedians(L2, k-((int)L1.size())-1);
    }

}

int main()
{
    int values[] = {2, 3, 5, 4, 1, 12, 11, 13, 16, 7, 8, 6, 10, 9, 17, 15, 19, 20, 18, 23, 21, 22, 25, 24, 14};

    vector<int> vec(values, values + 25);

    cout<<"The given array is : "<<endl;
    for (int i = 0; i < vec.size(); i++) {
        cout<<vec[i]<<" ";
    }

    selectionByMedianOfMedians(vec, 8);

    return 0;
}

下面是一个随机化快速选择的c++实现。这个想法是随机选择一个主元。为了实现随机分区，我们使用一个随机函数rand()来生成l和r之间的索引，将随机生成索引处的元素与最后一个元素交换，最后调用以最后一个元素为枢轴的标准分区过程。

#include<iostream>
#include<climits>
#include<cstdlib>
using namespace std;

int randomPartition(int arr[], int l, int r);

// This function returns k'th smallest element in arr[l..r] using
// QuickSort based method.  ASSUMPTION: ALL ELEMENTS IN ARR[] ARE DISTINCT
int kthSmallest(int arr[], int l, int r, int k)
{
    // If k is smaller than number of elements in array
    if (k > 0 && k <= r - l + 1)
    {
        // Partition the array around a random element and
        // get position of pivot element in sorted array
        int pos = randomPartition(arr, l, r);

        // If position is same as k
        if (pos-l == k-1)
            return arr[pos];
        if (pos-l > k-1)  // If position is more, recur for left subarray
            return kthSmallest(arr, l, pos-1, k);

        // Else recur for right subarray
        return kthSmallest(arr, pos+1, r, k-pos+l-1);
    }

    // If k is more than number of elements in array
    return INT_MAX;
}

void swap(int *a, int *b)
{
    int temp = *a;
    *a = *b;
    *b = temp;
}

// Standard partition process of QuickSort().  It considers the last
// element as pivot and moves all smaller element to left of it and
// greater elements to right. This function is used by randomPartition()
int partition(int arr[], int l, int r)
{
    int x = arr[r], i = l;
    for (int j = l; j <= r - 1; j++)
    {
        if (arr[j] <= x) //arr[i] is bigger than arr[j] so swap them
        {
            swap(&arr[i], &arr[j]);
            i++;
        }
    }
    swap(&arr[i], &arr[r]); // swap the pivot
    return i;
}

// Picks a random pivot element between l and r and partitions
// arr[l..r] around the randomly picked element using partition()
int randomPartition(int arr[], int l, int r)
{
    int n = r-l+1;
    int pivot = rand() % n;
    swap(&arr[l + pivot], &arr[r]);
    return partition(arr, l, r);
}

// Driver program to test above methods
int main()
{
    int arr[] = {12, 3, 5, 7, 4, 19, 26};
    int n = sizeof(arr)/sizeof(arr[0]), k = 3;
    cout << "K'th smallest element is " << kthSmallest(arr, 0, n-1, k);
    return 0;
}

上述解的最坏情况时间复杂度仍为O(n2)。在最坏的情况下，随机函数可能总是选择一个角元素。上述随机化QuickSelect的期望时间复杂度为Θ(n)

如果你想要一个真正的O(n)算法，而不是O(kn)或类似的算法，那么你应该使用快速选择(它基本上是快速排序，你会丢弃你不感兴趣的分区)。我的教授写了一篇很棒的文章，包括运行时分析:(参考)

QuickSelect算法可以快速找到包含n个元素的无序数组中的第k个最小元素。这是一个随机算法，所以我们计算最坏情况下的预期运行时间。

这是算法。

QuickSelect(A, k)
  let r be chosen uniformly at random in the range 1 to length(A)
  let pivot = A[r]
  let A1, A2 be new arrays
  # split into a pile A1 of small elements and A2 of big elements
  for i = 1 to n
    if A[i] < pivot then
      append A[i] to A1
    else if A[i] > pivot then
      append A[i] to A2
    else
      # do nothing
  end for
  if k <= length(A1):
    # it's in the pile of small elements
    return QuickSelect(A1, k)
  else if k > length(A) - length(A2)
    # it's in the pile of big elements
    return QuickSelect(A2, k - (length(A) - length(A2))
  else
    # it's equal to the pivot
    return pivot

这个算法的运行时间是多少?如果对手为我们抛硬币，我们可能会发现主元总是最大的元素，k总是1，给出的运行时间为

T(n) = Theta(n) + T(n-1) = Theta(n2)

但如果选择确实是随机的，则预期运行时间由

T(n) <= Theta(n) + (1/n) ∑i=1 to nT(max(i, n-i-1))

我们做了一个不完全合理的假设递归总是落在A1或A2中较大的那个。

让我们猜测对于某个a T(n) <= an，然后我们得到

T(n) 
 <= cn + (1/n) ∑i=1 to nT(max(i-1, n-i))
 = cn + (1/n) ∑i=1 to floor(n/2) T(n-i) + (1/n) ∑i=floor(n/2)+1 to n T(i)
 <= cn + 2 (1/n) ∑i=floor(n/2) to n T(i)
 <= cn + 2 (1/n) ∑i=floor(n/2) to n ai

现在我们要用加号右边这个可怕的和来吸收左边的cn。如果我们将其限定为2(1/n)∑i=n/2到n an，我们大致得到2(1/n)(n/2)an = an。但是这个太大了，没有多余的空间来挤进一个cn。让我们用等差级数公式展开和:

∑i=floor(n/2) to n i  
 = ∑i=1 to n i - ∑i=1 to floor(n/2) i  
 = n(n+1)/2 - floor(n/2)(floor(n/2)+1)/2  
 <= n2/2 - (n/4)2/2  
 = (15/32)n2

我们利用n“足够大”的优势，用更干净(更小)的n/4替换丑陋的地板(n/2)因子。现在我们可以继续

cn + 2 (1/n) ∑i=floor(n/2) to n ai,
 <= cn + (2a/n) (15/32) n2
 = n (c + (15/16)a)
 <= an

提供了> 16c。

得到T(n) = O(n)显然是(n)所以我们得到T(n) = (n)

还有Wirth的选择算法，它的实现比QuickSelect简单。Wirth的选择算法比QuickSelect慢，但经过一些改进，它变得更快。

更详细地说。使用Vladimir Zabrodsky的MODIFIND优化和3中位数的枢轴选择，并注意算法划分部分的最后步骤，我提出了以下算法(想象一下，命名为“LefSelect”):

#define F_SWAP(a,b) { float temp=(a);(a)=(b);(b)=temp; }

# Note: The code needs more than 2 elements to work
float lefselect(float a[], const int n, const int k) {
    int l=0, m = n-1, i=l, j=m;
    float x;

    while (l<m) {
        if( a[k] < a[i] ) F_SWAP(a[i],a[k]);
        if( a[j] < a[i] ) F_SWAP(a[i],a[j]);
        if( a[j] < a[k] ) F_SWAP(a[k],a[j]);

        x=a[k];
        while (j>k & i<k) {
            do i++; while (a[i]<x);
            do j--; while (a[j]>x);

            F_SWAP(a[i],a[j]);
        }
        i++; j--;

        if (j<k) {
            while (a[i]<x) i++;
            l=i; j=m;
        }
        if (k<i) {
            while (x<a[j]) j--;
            m=j; i=l;
        }
    }
    return a[k];
}

在我这里做的基准测试中，LefSelect比QuickSelect快20-30%。

你可以用O(n + kn) = O(n)(对于常数k)表示时间，用O(k)表示空间，通过跟踪你见过的最大的k个元素。

对于数组中的每个元素，您可以扫描k个最大的元素列表，并将最小的元素替换为更大的新元素。

Warren的优先级堆解决方案更简洁。