forked from algorithm019/algorithm019
-
Notifications
You must be signed in to change notification settings - Fork 0
Expand file tree
/
Copy pathMergeSort.java
More file actions
61 lines (49 loc) · 1.83 KB
/
MergeSort.java
File metadata and controls
61 lines (49 loc) · 1.83 KB
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
import java.util.Arrays;
/**
* Description: 归并排序
*
* 采用分治法(Divide and Conquer)的一个非常典型的应用。
* 将已有序的子序列合并,得到完全有序的序列;
* 即先使每个子序列有序,再使子序列段间有序。
* 若将两个有序表合并成一个有序表,称为2路归并
*
* 时间复杂度:O(nlogn)
* Date: 2020-12-26
* Time: 11:53 AM
*/
public class MergeSort {
public static void mergeSort(int[] arr,int left,int right) {
if (right <= left) {
return;
}
int mid = (left + right) >> 1;
mergeSort(arr, left, mid);
mergeSort(arr, mid + 1, right);
merge(arr, left, mid, right);
}
public static void merge(int[] arr, int left, int mid, int right) {
int[] temp = new int[right - left + 1];
int i = left, j = mid + 1, k = 0;
//左右指针同时移动,将两边数组按由大到小的顺序进行合并
while (i <= mid && j <= right) {
temp[k++] = arr[i] <= arr[j] ? arr[i++] : arr[j++];
}
//如果左边的指针没移动到mid处,说明左边数组还有元素没有处理完毕,直接按顺序往后累加即可
while (i <= mid) {
temp[k++] = arr[i++];
}
//如果右边的指针没移动到mid处,说明右边数组还有元素没有处理完毕,直接按顺序往后累加即可
while (j <= right) {
temp[k++] = arr[j++];
}
for (int p = 0; p < temp.length; p++) {
arr[left + p] = temp[p];
}
// System.arraycopy(temp,0,arr,left,temp.length);
}
public static void main(String[] args) {
int[] arr = new int[]{1, 4, 3, 7, 2, 10, 5, 21, 6};
MergeSort.mergeSort(arr,0,8);
System.out.println(Arrays.toString(arr));
}
}