/**
線性時間復(fù)雜度求數(shù)組中第K小數(shù)
** author :liuzhiwei
** data :2011-08-07
**/
#include "iostream"
using namespace std;
//基于快速排序思想，求數(shù)組a中第k小的數(shù)，low和high分別為數(shù)組的起始和結(jié)束位置
//時間復(fù)雜度為o(n)，n為數(shù)組的長度
//1<=k<=n
//如果存在，返回第k小數(shù)的下標(biāo)，否則返回-1
int selectk(int a[], int low, int high, int k)
{
if(k <= 0)
return -1;
if(k > high - low + 1)
return -1;
int pivot = low + rand()%(high - low + 1); //隨即選擇一個支點
swap(a[low], a[pivot]);
int m = low;
int count = 1;
//一趟遍歷，把較小的數(shù)放到數(shù)組的左邊
for(int i = low + 1; i <= high; ++i)
{
if(a[i]<a[low])
{
swap(a[++m], a[i]);
count++; //比支點小的數(shù)的個數(shù)為count-1
}
}
swap(a[m], a[low]); //將支點放在左、右兩部分的分界處
if(k < count)
{
return selectk(a, low, m - 1, k);
}
else if( k > count)
{
return selectk(a, m + 1, high, k - count);
}
else
{
return m;
}
}
int main(void)
{
int a[] = {5, 15, 5, 7, 9, 17,100, 3, 12, 10, 19, 18, 16, 10, 1000,1,1,1,1,1,1,1,1};
int r = selectk(a, 0, sizeof(a) /sizeof(int) - 1, 23);
cout<<(r == -1 ? r : a[r])<<endl;
system("pause");
return 0;
}
