【编程题目】查找最小的 k 个元素

/*
5.查找最小的 k 个元素（数组）
题目：输入 n 个整数，输出其中最小的 k 个。
例如输入 1，2，3，4，5，6，7 和 8 这 8 个数字，则最小的 4 个数字为 1，2，3 和 4。
*/

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

bool compare(int a, int b) //降序排列
{
    return a > b;
}

//找到 一共有 n个 元素的 数组S 中 第k小 的数字  整个算法中，对除以5后余下的数字都做了特殊处理
int Select(int * S, int k, int n)
{
    if (n < 5) //对少于5个的做特殊处理
    {
        sort(S, S + n);
        return S[k - 1];
    }
    int subn = n/5 + ((n % 5 == 0) ? 0 : 1);
    int subnn = n/5;
    int (* subS)[5] = new int[subn][5];
    for (int i = 0; i < subnn; i++)
    {
        for (int j = 0; j < 5; j++)
        {
            subS[i][j] = S[i * 5 + j]; 
        }
        sort(subS[i], subS[i] + 5, compare); //5个一组,每组从大到小排序
    }
    for (int j = 0; j < n % 5; j++)
    {
        subS[subn - 1][j] = S[subn * 5 + j - 5];
    }
    sort(subS[subn - 1], subS[subn - 1] + n % 5, compare); 

    int * M = new int [subn];
    for (int i = 0; i < subn; i++)
    {
        M[i] = subS[i][2]; //M 中存储每组数字的中位数
    }

    int Mn = subnn;
    int m = Select(M, Mn/2 + (Mn % 2 == 0) ? 0 : 1, Mn);
    delete [] M;

    int * S1 = new int [n]; //存放小于等于m的数字
    int * S2 = new int [n];
    int S1n = 0; //记录有多少小于等于m的数字
    int S2n = 0;

    //找到相应的S1 与 S2中的元素
    for (int i = 0; i < subnn; i++)
    {
        if (subS[i][2] <= m)
        {
            for (int j = 0; j < 5; j++)
            {
                if (j < 2)
                {
                    if (subS[i][j] <= m)
                    {
                        S1[S1n++] = subS[i][j];
                    }
                    else
                    {
                        S2[S2n++] = subS[i][j];
                    }
                }
                else
                {
                    S1[S1n++] = subS[i][j];
                }
            }
        }
        else
        {
            for (int j = 0; j < 5; j++)
            {
                if (j > 2)
                {
                    if (subS[i][j] <= m)
                    {
                        S1[S1n++] = subS[i][j];
                    }
                    else
                    {
                        S2[S2n++] = subS[i][j];
                    }
                }
                else
                {
                    S2[S2n++] = subS[i][j];
                }
            }
        }
    }
    if (subnn != subn) //多余的数字特别处理
    {
        for (int j = 0; j < n % 5; j++)
        {
            if (subS[subn - 1][j] > m)
            {
                S2[S2n++] = subS[subn - 1][j];
            }
            else
            {
                S1[S1n++] = subS[subn - 1][j];
            }
        }
    }

    if (k == S1n)
    {
        delete [] S1;
        delete [] S2;
        return m;
    }
    else if (k < S1n)
    {
        return Select(S1, k, S1n);
    }
    else
    {
        return Select(S2, k - S1n, S2n);
    }

    delete [] S1;
    delete [] S2;
}

int main()
{
    int a[8] = {1,2,3,4,5,6,7,8};
    int m = Select(a, 7, 8);

    return 0;
}

public static void FindKMin(int[] sort, int k)
{
    int[] heap = sort;
    int rootIndex = k / 2 - 1;
    while (rootIndex >= 0)
    {
        reheap(heap, rootIndex, k - 1);
        rootIndex--;
    }
 
    for (int i = k, len=heap.Length; i < len; i++)
    {
        if (heap[i]<heap[0])
        {
            heap[0] = heap[i];
            reheap(heap, 0, k - 1);
        }
    }
 
    Console.WriteLine("The {0} min element =",k);
    for (int i = 0; i < k; i++)
    {
        Console.Write(heap[i] + " ");
    }
}
 
private static void reheap(int[] heap, int rootIndex, int lastInddex)
{
    int orphan = heap[rootIndex];
    bool done = false;
    int leftIndex = rootIndex * 2 + 1;
    while (!done && leftIndex <= lastInddex)
    {
        int largerIndex = leftIndex;
        if (leftIndex+1 <= lastInddex)
        {
            int rightIndex = leftIndex + 1;
            if (heap[rightIndex] > heap[leftIndex])
            {
                largerIndex = rightIndex;
            }
        }
 
        if (orphan < heap[largerIndex])
        {
            heap[rootIndex] = heap[largerIndex];
            rootIndex = largerIndex;
            leftIndex = rootIndex * 2 + 1;
        }
        else
        {
            done = true;
        }
    }
 
    heap[rootIndex] = orphan;
}