算法导论-9.3-7
【转】
题目:
给出一个O(n)时间的算法,在给定一个有n个不同数字的集合S以及一个正整数k<=n后,它能确定出S中最接近其中位数的k个数。
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算法思想:
1.找到数组a中第n/2小的数median;
2.对a中非median数进行|a[i] - median|,得到一个大小为n - 1的数组distance;
3.寻找distance中第k小的数值;
4.对distance进行一次遍历,找到小于等于k的数,从而对应得到数组a中的k个数。
上述每一步的时间复杂度都为O(n),因而最后总的时间复杂度为O(n).
#include <iostream>
#include <time.h>
using namespace std;
//随机化分割
int randomized_partition(int* a, int p, int r);
int randomized_select(int* a, int p, int r, int i);
int main()
{
const int LEN = 12;
int arr[LEN] = { 5, 7, 10, 3, 6, 2, 8, 9, 4, 1, 11 , 12};
int median = randomized_select(arr, 0, LEN - 1, (LEN - 1)/2);
int distance[LEN - 1];
int distance_cpy[LEN - 1];
int temp = 0;
int middle;
for(int i = 0; i < LEN; i++)
{
if(arr[i] != median)
distance[temp++] = abs(arr[i] - median);
else
middle = i;
}
for(int i = 0; i < LEN - 1; i++)
distance_cpy[i] = distance[i];
int k;
while(true)
{
cout<<"输入k:"<<endl;
cin>>k;
cout<<"k邻近值为:"<<endl;
int kth_value = randomized_select(distance_cpy, 0, LEN - 2, k - 1);
int* k_arr = new int[k];
int j = 0;
//这里要注意,比如中间值为6,那么k = 3,有5,7明显符合要求,他们最近,4,8也符合要求,那么在这选择谁
if(k%2)
{
bool flag = true;
for(int i = 0; i < LEN - 1; i++)
{
if(distance[i] < kth_value)
{
if(i < middle)
k_arr[j++] = median - distance[i];
else
k_arr[j++] = median + distance[i];
}
else if(distance[i] == kth_value && flag)
{
if(i < middle)
k_arr[j++] = median - distance[i];
else
k_arr[j++] = median + distance[i];
flag = false;
}
}
}
else
{
for(int i = 0; i < LEN - 1; i++)
{
if(distance[i] <= kth_value)
{
if(i < middle)
k_arr[j++] = median - distance[i];
else
k_arr[j++] = median + distance[i];
}
}
}
for(int i = 0; i < k; i++)
cout<<k_arr[i]<<"\t";
cout<<endl;
delete[] k_arr;
}
}
//下标为[p, r]之间的元素
int randomized_partition(int* a, int p, int r)
{
srand(time(NULL));
int q = rand()%(r - p + 1) + p;
int temp = a[q];
a[q] = a[r];
a[r] = temp;
int j = p;
for(int i = p; i < r; i++)
{
if(a[i] < a[r])
{
if(i != j)
{
int temp2 = a[i];
a[i] = a[j];
a[j] = temp2;
}
j++;
}
}
temp = a[j];
a[j] = a[r];
a[r] = temp;
return j;
}
//迭代版本
int randomized_select(int* a, int p, int r, int i)
{
int q = randomized_partition(a, p, r);
while(p != r)
{
if(i == q)
return a[q];
else if(i < q)
{
r = q - 1;
q = randomized_partition(a, p, r);
}
else
{
p = q + 1;
q = randomized_partition(a, p, r);
}
}
return a[p];
}