顺序统计

一.最大值与最小值

 1 #include<iostream>
 2 using namespace std;
 3
 4 template<typename T>
 5 void MaxMin(T A[], int length, T &max, T &min)
 6 {
 7     if (A == NULL || length <= 0)
 8     {
 9         return;
10     }
11     int i;
12     if (length % 2 == 0)
13     {
14         i = 2;
15         if (A[0] > A[1])
16         {
17             max = A[0];
18             min = A[1];
19         }
20         else
21         {
22             max = A[1];
23             min = A[0];
24         }
25     }
26     else
27     {
28         max = A[0];
29         min = A[0];
30         i = 1;
31     }
32
33     for (; i < length; i += 2)
34     {
35         T cmax, cmin;
36         if (A[i] > A[i + 1])
37         {
38             cmax = A[i];
39             cmin = A[i + 1];
40         }
41         else
42         {
43             cmax = A[i + 1];
44             cmin = A[i];
45         }
46
47         if (max < cmax)
48         {
49             max = cmax;
50         }
51         if (min > cmin)
52         {
53             min = cmin;
54         }
55     }
56 }
57
58 int main()
59 {
60     int A[] = { 111, 2, 37, 3, 4, 100, 0, 600, 20, 10, -11};
61     int max, min;
62     MaxMin(A, 11, max, min);
63     cout << max << " " << min << endl;
64     return 0;
65 }

最多3*(n / 2)次比较

二.在最坏情况下 找到n个元素第二小

http://blog.csdn.net/mishifangxiangdefeng/article/details/7983809

三.线性选择算法

 1 #include<iostream>
 2 #include<stdlib.h>
 3 using namespace std;
 4
 5 template <typename T>
 6 int Partition(T A[], int p, int r)
 7 {
 8     T x = A[r];
 9     int i = p - 1;
10     for (int j = p; j <= r - 1; j++)
11     {
12         if (A[j] <= x)
13         {
14             i++;
15             T temp = A[j];
16             A[j] = A[i];
17             A[i] = temp;
18         }
19     }
20     T temp = A[i + 1];
21     A[i + 1] = A[r];
22     A[r] = temp;
23     return i + 1;
24 }
25
26 template <typename T>
27 int RandomizedPartition(T A[], int p, int r)
28 {
29     int i = rand() % (r - p + 1) + p;
30     T temp = A[r];
31     A[r] = A[i];
32     A[i] = temp;
33     return Partition(A, p, r);
34 }
35
36 template <typename T>
37 T RandomizedSelect(T A[], int p, int r, int i)
38 {
39     if (A != NULL && p >= 0 && r >= 0 && r >= p)
40     {
41         if (r - p + 1 < i)
42         {
43             return NULL;
44         }
45         else
46         {
47             if (r == p)
48             {
49                 return A[p];
50             }
51             int q = RandomizedPartition(A, p, r);
52             int k = q - p + 1;
53             if (i == k)
54             {
55                 return A[q];
56             }
57             else if (i < k)
58             {
59                 return RandomizedSelect(A, p, q - 1, i);
60             }
61             else
62             {
63                 return RandomizedSelect(A, q + 1, r, i - k);
64             }
65         }
66     }
67 }
68
69 int main()
70 {
71     double A[] = { 1.1, -2.1, 3, 3, 4, -10, 4, 2.6 ,20, 10};
72     cout << RandomizedSelect(A, 0, 9, 2) << endl; //-2.1
73     return 0;
74 }

四.最坏情况为线性的选择

1.将输入数组的n个元素划分为n/5组，每组5个元素，剩余的一组由n mod 5个元素组成

2.寻找n/5组，每组的中位数

3.对n/5个中位数用select找出其中位数x

4.用中位数x对输入的数进行划分