[算法设计与分析]4.3.5非等分分治(一组数的第二小+一组数的第k小)

#include<stdio.h>
#include<iostream>
#include<cstring>

using namespace std;

const int N = 7;

int SecondSmall(int a[], int n);
void FindTwo(int a[], int i, int j, int &fmin1, int &fmin2);
int FindK(int a[], int n, int k);
void Swap(int &left, int &right);
int Select(int a[], int left, int right, int k);

int main ()
{
    int a[N] = {2, 4, 1, 6, 5, 9, 0};
    //cout << SecondSmall(a, N) << " is the second small number" << endl;

    int k = 3;
    cout << FindK(a, N, k) << " is the " << k << " small number" << endl;
}

//利用查找最小的两个数 才可以将原问题进行分解
int SecondSmall(int a[], int n)
{
    int min1, min2;
    FindTwo(a, 0, n - 1, min1, min2);
    return min2;
}

void FindTwo(int a[], int i, int j, int &fmin1, int &fmin2)
{
    int lmin1, lmin2, rmin1, rmin2;
    int mid;

    //递归出口
    if(i == j)
        fmin2 = fmin1 = a[i];
    else if(i == j - 1)
    {
        fmin2 =