例如{2，4，3，4，7}中，第一大的数是7，位置在4。第二大、第三大的数都是4，位置在1、3随便输出哪一个均可。函数接口为：int find_orderk(const int* narry,const int n,const int k)
要求算法复杂度不能是O(n^2）
//快速排序
#include
usingnamespacestd;
intPartition (int*L,intlow,int high)
{
inttemp = L[low];
intpt = L[low];
while (low < high){while (low < high && L[high] >= pt)
–high;
L[low] = L[high];
while (low < high && L[low] <= pt)++low;L[low] = temp;}L[low] = temp;returnlow;}voidQSort (int*L,intlow,int high){if (low < high){intpl = Partition (L,low,high);QSort (L,low,pl - 1);QSort (L,pl + 1,high);}}intmain (){intnarry[100],addr[100];intsum = 1,t;cout << "Input number:" << endl;cin >> t;
while (t != -1)
{
narry[sum] = t;
addr[sum – 1] = t;
sum++;
cin >> t;
}
sum -= 1;
QSort (narry,1,sum);
for (int i = 1; i <= sum;i++)cout << narry[i] << 't';cout << endl;intk;cout << "Please input place you want:" << endl;cin >> k;
intaa = 1;
intkk = 0;
for (;;)
{
if (aa == k)
break;
if (narry[kk] != narry[kk + 1])
{
aa += 1;
kk++;
}
}
cout << "The NO." << k << "number is:" << narry[sum - kk] << endl;cout << "And it's place is:" ;for (i = 0;i < sum;i++){if (addr[i] == narry[sum - kk])cout << i << 't';}return0;}
