c++ - quickselect 算法因重复元素而失败

Question

我已经实现了快速选择算法。

我有一个问题，当我在数组中使用重复项时，我的算法会陷入无限循环...

你能帮我让它工作吗？

预期复杂度为 O(n)，最坏情况为 O(n^2)？

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

int rand_partition(vector<int> &a, int left, int right) { 
    int pivotIndex = left + (rand() % (right - left)); 
    //int m = left + (right - left) / 2; //... to test the algo...no rand at this point 
    int pivot = a[pivotIndex]; 
    int i = left; 
    int j = right; 

    do { 
        while (a[i] < pivot) i++; // find left element > pivot 
        while (a[j] > pivot) j--; // find right element < pivot 

        // if i and j not already overlapped, we can swap 
        if (i < j) { 
            swap(a[i], a[j]); 
        } 
    } while (i < j); 

    return i; 
} 

// Returns the n-th smallest element of list within left..right inclusive (i.e. n is zero-based). 
int quick_select(vector<int> &a, int left, int right, int n) { 
    if (left == right) {        // If the list contains only one element 
        return a[left];  // Return that element 
    } 

    int pivotIndex = rand_partition(a, left, right); 

    // The pivot is in its final sorted position 
    if (n == pivotIndex) { 
        return a[n]; 
    } 
    else if (n < pivotIndex) { 
        return quick_select(a, left, pivotIndex - 1, n); 
    } 
    else { 
        return quick_select(a, pivotIndex + 1, right, n); 
    } 
} 

int main() { 

    vector<int> vec= {1, 0, 3, 5, 0, 8, 6, 0, 9, 0}; 

    cout << quick_select(vec, 0, vec.size() - 1, 5) << endl; 

    return 0; 
}

Answer 1

您的代码中存在多个问题。

• 首先，在函数 quick_select() 中, 你正在比较 pivotIndex与 n直接地。自 left不总是 0，你应该比较 n左边部分的长度等于pivotIndex - left + 1 .
• 何时n > length , 你只需调用 quick_select(a, pivotIndex + 1, right, n)递归地，这时，就是整个 vector 的第N个元素在它的右半部分，就是右半部分的第(N - (pivotIndex - left + 1) )-th 个元素 vector 。代码应该是quick_select(a, pivotIndex + 1, right, n - (pivotIndex - left + 1) ) (n 是基于 ONE 的)。
• 您似乎在使用 Hoare 的分区算法并且实现不正确。即使它有效，当 HOARE-PARTITION 终止时，它返回一个值 j 使得 A[p...j] ≤ A[j+1...r] , 但我们想要 A[p...j-1] ≤ A[j] ≤ A[j+1...r]在quick_select() .所以我使用 rand_partition()基于我在 another post 上写的 Lomuto 分区算法

这是固定的 quick_select()它返回 vector 的第 N 个(请注意 n 是基于 ONE 的)最小元素:

int quick_select(vector<int> &a, int left, int right, int n)
{
    if ( left == right ) 
        return a[left];
    int pivotIndex = partition(a, left, right);

    int length = pivotIndex - left + 1;
    if ( length == n) 
        return a[pivotIndex];
    else if ( n < length ) 
        return quick_select(a, left, pivotIndex - 1, n);
    else 
        return quick_select(a, pivotIndex + 1, right, n - length);
}

和this是rand_partition() :

int rand_partition(vector<int> &arr, int start, int end)
{
    int pivot_index = start + rand() % (end - start + 1);
    int pivot = arr[pivot_index];

    swap(arr[pivot_index], arr[end]); // swap random pivot to end.
    pivot_index = end;
    int i = start -1;

    for(int j = start; j <= end - 1; j++)
    {
        if(arr[j] <= pivot)
        {
            i++;
            swap(arr[i], arr[j]);
        }
    }
    swap(arr[i + 1], arr[pivot_index]); // swap back the pivot

    return i + 1;
}

调用srand()首先初始化随机数生成器，以便在调用 rand() 时可以获得类似随机数的数字。 .
测试上述功能的驱动程序:

int main()
{
    int A1[] = {1, 0, 3, 5, 0, 8, 6, 0, 9, 0};
    vector<int> a(A1, A1 + 10);
    cout << "6st order element " << quick_select(a, 0, 9, 6) << endl;
    vector<int> b(A1, A1 + 10); // note that the vector is modified by quick_select()
    cout << "7nd order element " << quick_select(b, 0, 9, 7) << endl;
    vector<int> c(A1, A1 + 10);
    cout << "8rd order element " << quick_select(c, 0, 9, 8) << endl;
    vector<int> d(A1, A1 + 10);
    cout << "9th order element " << quick_select(d, 0, 9, 9) << endl;
    vector<int> e(A1, A1 + 10);
    cout << "10th order element " << quick_select(e, 0, 9, 10) << endl;
}