CLRS 9.3.7----最接近中位数的k个数 - liken的专栏 - 博客频道 - CSDN.NET

算法导论第九章9.3.7----最接近中位数的k个数 - liken的专栏 - 博客频道 - CSDN.NET
Describe an O(n) algorithm that, given a set S of n distinct numbers
and a positive integer k ≤ n, determines the k numbers in S that are closest to the
median of S .

用O（n）找到中位数，然后每个数都减去中位数并且取绝对值，得到一个新数组，然后，找出新数组里的最小的k个数，也是花费O（n），所以最后的时间复杂度还是O（n）

Solution: 
Note that the k numbers in S that are closest to the median are among the k smallest numbers on either side of the partition (around the median).
注意到数组中最接近中位数的k个数字是两边分区中最小的k个数。
We first use linear time selection to find the median. 
首先使用线性选择算法找到中位数。
Then we compute an array B of absolute differences between the ith element in S and the median, B[i] = |S [i] − median|.
之后计算数字B，该数组记录数组S中每个元素到中位数距离的绝对值。B[i] = |S [i] − median|.
The k elements closest to the median are the k smallest elements in B. 
最接近中位数的k个元素是B中最小的k个元素。
We use linear time selection to find the k-th smallest element in B. 
使用线性选择算法找到数组B中第k个最小元素。
The first k elements in the resulting partition of B correspond to the k numbers in S closest to the median. 
数组B经过选择算法处理后，在结果划分的前k个元素就对应数组S中距离中位数最近的k个数。
The algorithm takes O(n) time as we use linear time selection exactly two times and traverse the n numbers in S once.

使用两次线性时间选择算法找中位数和第k个数，一次遍历数组S中的n个数，该算法时间复杂度为O(n)。

1 get the median of S
2 compute the distence of S to the median, store in B
3 get the kth order statistics in B. the first k elements in partition of B is the output

1. We find the median of the array in linear time
2. We find the distance of each other element to the median in linear time
3. We find the k-th order statistic of the distance, again, in linear time
4. We select only the elements that have distance lower than or equal to the k-th order statistic

1: procedure k_Closest(S, k)  //S: a set of n numbers and k: an integer
2: Output = nothing;
3: m = Select(S, n,n/2)                       //O(n)
4: for all  s in S and s != m               //O(n)
5: s.distance = |m − s|
6: end for
7: md = Select(S.distance, k)        //O(n)
8: for all s in S
9: if s.distance <= md.distance then   //O(n)
10: Output = Output + s
11: end if
12: end for
13: return Output
14: end procedure

1. /******寻找最接近中位数的k个数*******/
2. //length：数组的长度
3. void FindKNums(int a[], int length, int k)
4. {
5.     if(k > length)
6.     {
7.         return;
8.     }
9. 
   
10.     //step1:求出数组的中位数的值O(n)
11.     int mid = Select(a, 0, length - 1, length / 2);
12.     
13.     cout << "mid = " << mid << endl;
14. 
    
15.     //step2:计算数组每个数与中位数差的绝对值，存于另一个数组B中O(n)
16.     int *b = (int *)malloc(sizeof(int) * length);
17.     if(b == NULL)
18.     {
19.         return;
20.     }
21.     
22.     for(int i = 0; i < length; i++)
23.     {
24.         b[i] = (int)fabs(double(a[i] - mid));
25.     }
26. 
    
27.     //step3:求出数组B中第k小的数ret O(n)
28.     int ret = Select(b, 0, length - 1, k);
29. 
    
30.     cout << "ret = " << ret << endl;
31. 
    
32.     //step4:计算数组S中与中位数的差的绝对值小于ret的数并输出O(n)
33.     for(int i = 0; i < length; i++)
34.     {
35.         int tmp = (int)fabs(double(a[i] - mid));
36.         if(tmp <= ret)
37.         {
38.             cout << a[i] << " ";
39.         }
40.     }
41.     cout << endl;
42. 
    
43. }

http://www.cnblogs.com/shuaiwhu/archive/2011/04/13/2065064.html
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