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Reverse Pairs

时间:2016-07-10 13:51:31      阅读:116      评论:0      收藏:0      [点我收藏+]

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For an array A, if i < j, and A [i] > A [j], called (A [i], A [j]) is a reverse pair.
return total of reverse pairs in A.

Example

Given A = [2, 4, 1, 3, 5] , (2, 1), (4, 1), (4, 3) are reverse pairs. return 3

 1 public class Solution {
 2     /**
 3      * @param A an array
 4      * @return total of reverse pairs
 5      */
 6     public long reversePairs(int[] A) {
 7         if (A == null || A.length == 0) {
 8             return 0;
 9         }
10         int[] temp = new int[A.length];
11         return mergeSort(A, temp, 0, A.length - 1);
12     }
13     
14     private long mergeSort(int[] A, int[] temp, int left, int right) {
15         if (left >= right) {
16             return 0;
17         }
18         
19         int sum = 0;
20         int mid = left + (right - left) / 2;
21         sum += mergeSort(A, temp, left, mid);
22         sum += mergeSort(A, temp, mid + 1, right);
23         sum += merge(A, temp, left, mid, right);
24         return sum;
25     }
26     
27     private long merge(int[] A, int[] temp, int left, int mid, int right) {
28         int i = left, j = mid + 1;
29         int sum = 0;
30         for (int k = 0; k < right - left + 1; k++) {
31             if (i <= mid && (j > right || A[i] <= A[j])) {
32                 temp[k] = A[i];
33                 i++;
34             } else {
35                 temp[k] = A[j];
36                 if (i <= mid) {
37                 sum += mid - i + 1;
38                 }
39                 j++;
40             }
41         }
42         for (int k = 0; k < right - left + 1; k++) {
43             A[left + k] = temp[k];
44         }
45         return sum;
46     }
47 }

 

Reverse Pairs

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原文地址:http://www.cnblogs.com/FLAGyuri/p/5657442.html

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