剑指offer - 面试题30 找出数组最小的k个数

//方法1 : 通过类似快排的方法找到排在第k+1位置的数,并且左边都是比他小的,因此桌面的书就是k个最小的数
#include
#include
using namespace std;
void Swap(int &a, int &b)
{
    int temp = a;
    a = b;
    b = temp;
}
//和快排类似, 找一个枢纽元,把小于枢纽元的数放在枢纽元左边, 返回枢纽元的位置
int Partion(int *nArray, int nBegin, int nEnd)
{
    if(nArray == NULL|| nBegin <0 || nEnd<0)
        throw exception("ERROR");
    int nPivot = nArray[nBegin];  //枢纽元
    int nLeft = nBegin;
    int nRight = nEnd;
    while(nLeft != nRight)
    {
        while(nArray[nRight]>=nPivot && nLeft < nRight)
            --nRight;
        while(nArray[nLeft]<=nPivot && nLeft < nRight)
            ++nLeft;
        if(nLeftreturn nLeft;
}
void GetLeastNum(int *nArray, int nLength, int k)
{
    if(nArray == NULL || nLength <= 0)
        throw exception("ERROR");
    int nBegin = 0;
    int nEnd = nLength-1;
    //找出排在第k+1个位置的那个数的值.
    int nNum = Partion(nArray, nBegin, nEnd);
    while(nNum != k+1)
    {
        if(nNum > k+1)
        {
            nNum = Partion(nArray, nBegin, nNum-1);
        }
        else
        {
            nNum = Partion(nArray, nNum+1, nEnd);
        }
    }
    for(int i = 0; i < k;++i)
    {
        cout<", ";
    }
    cout<int main()
{
    int a[] = {1,2,3,4,5,6,7,8,9,0};
    GetLeastNum(a, 10, 5);
}