隨機播放演算法

      對於一個無序的數列，如何以線性時間選擇其中位元？快速排序可以，但時間複雜度為O(nlogn)，我們並不想得到整個數列的順序，只需要其中位元，即第n/2個數。因此，我們從快速排序演算法的思路得到隨機播放演算法，時間複雜度為o(n)

#include <iostream></p><p>using namespace std;</p><p>#include <stdlib.h><br />#include <conio.h><br />#include <time.h> </p><p>int Partition(int L[],int first,int last)<br />{<br />int left = first;<br />int right = last;<br />int pivot = L[first];</p><p>while (left < right)<br />{<br />while (left < right && L[right] >= pivot)right--;<br />L[left] = L[right];<br />while (left < right && L[left] < pivot) left++;<br />L[right] = L[left];<br />}<br />L[left] = pivot;<br />return left;<br />}<br />int Random_Partition(int L[],int first,int last)<br />{<br />int i;<br />int temp;</p><p>srand((unsigned)time(NULL));<br />i = rand() % (last - first + 1)+first;</p><p>temp = L[i];<br />L[i] = L[last];<br />L[last] = temp;<br />return Partition(L,first,last);<br />}<br />int RandomizeSelect(int L[],int first,int last,int mid_num)<br />{<br />int q,k;</p><p>if(first == last)return L[first];</p><p>q = Random_Partition(L,first,last);<br />k = q - first + 1;</p><p>if (mid_num == k)return L[q];<br />else if (mid_num < k)return RandomizeSelect(L,first,q - 1,mid_num);<br />else return RandomizeSelect(L,q + 1,last,mid_num - k);<br />}</p><p>void main()<br />{<br />int L[8] = {3,45,13,27,34,23,56,8};<br />int select;</p><p>select = RandomizeSelect(L,0,7,1);<br />cout<<select<<endl;<br />select = RandomizeSelect(L,0,7,2);<br />cout<<select<<endl;<br />select = RandomizeSelect(L,0,7,3);<br />cout<<select<<endl;<br />select = RandomizeSelect(L,0,7,4);<br />cout<<select<<endl;</p><p>select = RandomizeSelect(L,0,7,5);<br />cout<<select<<endl;<br />select = RandomizeSelect(L,0,7,6);<br />cout<<select<<endl;<br />select = RandomizeSelect(L,0,7,7);<br />cout<<select<<endl;<br />select = RandomizeSelect(L,0,7,8);<br />cout<<select<<endl;<br />}

 

參考：

[1] 演算法導論第九章（中位元和順序統計學）