逆数对 

 1 #include<iostream>
 2 using namespace std;
 3 
 4 int  InversePairsCore(int *data, int * copy, int start, int end)
 5 {
 6     if (start == end)
 7     {
 8         copy[start] = data[end];
 9         return 0;
10     }
11     int length = (end - start) / 2;
12 
13     int left = InversePairsCore(copy, data, start, start + length);
14     int right = InversePairsCore(copy, data, start + length+1,end);
15 
16     int i = start + length;
17     int j = end;
18     int indexcopy = end;
19     int count = 0;
20     while (i >= start&&j >= start + length + 1)
21     {
22         if (data[i] > data[j])
23         {
24             copy[indexcopy--] = data[i--];
25             count = j - start - length;
26         }
27         else
28         {
29             copy[indexcopy--] = data[j--];
30         }
31     }
32     for (; i >= start; i--)
33         copy[indexcopy--] = data[i];
34     for (; j >= start + length + 1; j--)
35         copy[indexcopy--] = data[j];
36     return left + right + count;
37 }
38 int inversepairs(int *data, int length)
39 {
40     if (data == NULL || length <= 0)
41         return 0;
42     int *copy = new int[length];
43     for (int i = 0; i < length; i++)
44         copy[i] = data[i];
45     int count = InversePairsCore(data, copy, 0, length - 1);
46     for (int i = 0; i < length; i++)
47         cout << copy[i] << " ";
48     cout << endl;
49     return count;
50 
51 }
52 
53 int main()
54 {
55     int data[] = { 7, 5, 6, 4 };
56     cout << inversepairs(data, 4);
57     cin.get();
58     return 0;
59 }

 

posted @ 2016-07-28 23:11  ranran1203  阅读(324)  评论(0)    收藏  举报