Summary: Merge Sort of Array && 求逆序对

常用算法：

 1 package ArrayMergeSort;
 2
 3 import java.util.Arrays;
 4
 5 public class Solution {
 6     public int[] mergeSort(int[] arr) {
 7         if (arr.length == 1) return arr;
 8         else {
 9             int[] arr1 = Arrays.copyOfRange(arr, 0, arr.length/2);
10             int[] arr2 = Arrays.copyOfRange(arr, arr.length/2, arr.length);
11             return merge(mergeSort(arr1), mergeSort(arr2));
12         }
13     }
14
15     public int[] merge(int[] arr1, int[] arr2) {
16         int len1 = arr1.length;
17         int len2 = arr2.length;
18         int[] res = new int[len1+len2];
19         int i = 0, j=0, cur=0;
20         while (i<len1 && j<len2) {
21             if (arr1[i] <= arr2[j]) {
22                 res[cur++] = arr1[i++];
23             }
24             else {
25                 res[cur++] = arr2[j++];
26             }
27         }
28         while (i<len1) {
29             res[cur++] = arr1[i++];
30         }
31         while (j<len2) {
32             res[cur++] = arr2[j++];
33         }
34         return res;
35     }
36
37
38
39     /**
40      * @param args
41      */
42     public static void main(String[] args) {
43         // TODO Auto-generated method stub
44         Solution sol = new Solution();
45         int[] arr = sol.mergeSort(new int[]{6,5,4,8,2,1});
46         System.out.println(Arrays.toString(arr));
47     }
48
49 }

在如上算法中只需稍作修改，加上一行代码，就可以求数组的逆序对

如数组 <2,3,8,6,1> 的逆序对为：<2,1> <3,1> <8,6> <8,1> <6,1> 共5个逆序对。

暴力法是O（N^2）

mergeSort可以O（NlogN）

定义一个static variable count, 然后在12行加入

 1 public int[] merge(int[] arr1, int[] arr2) {
 2         int len1 = arr1.length;
 3         int len2 = arr2.length;
 4         int[] res = new int[len1+len2];
 5         int i = 0, j=0, cur=0;
 6         while (i<len1 && j<len2) {
 7             if (arr1[i] <= arr2[j]) {
 8                 res[cur++] = arr1[i++];
 9             }
10             else { // arr1[i] > arr2[j];
11                 res[cur++] = arr2[j++];
12                 count += arr1.length - i;
13             }
14         }
15         while (i<len1) {
16             res[cur++] = arr1[i++];
17         }
18         while (j<len2) {
19             res[cur++] = arr2[j++];
20         }
21         return res;
22     }