剑指offer

//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<iostream>#include<vector>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(nLeft<nRight)        {            Swap(nArray[nLeft], nArray[nRight]);        }    }    Swap(nArray[nBegin], nArray[nLeft]);    return 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<<nArray[i]<<", ";    }    cout<<endl;}int main(){    int a[] = {1,2,3,4,5,6,7,8,9,0};    GetLeastNum(a, 10, 5);}