【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);elsemiddle = 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];elsek_arr[j++] = median + distance[i];}else if(distance[i] == kth_value && flag){if(i < middle)k_arr[j++] = median - distance[i];elsek_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];elsek_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];}
来源：https://www.cnblogs.com/longdouhzt/archive/2011/07/28/2119393.html