归并排序法02:归并排序法的优化
优化一:对有序数组排序的优化
在每次合并两个子数组前进行判断,如果左边数组的最大值都小于右边数组的最小值,就不用进行合并操作
对于完全有序的数组,递归树中每层都不用进行合并,最后一层叶子节点需要操作的次数是n,其上一层为1/2n,1/4n...最后总共需要操作的次数为2n
因此对于完全有序的数组,归并排序法的时间复杂度会降为O(n)
import java.util.Arrays;
public class Algorithm {
public static void main(String[] args) {
Integer[] arr = {6,6,3,2,4,7};
MergeSort.sortOptimised1(arr);
System.out.println(Arrays.toString(arr));
}
}
class MergeSort {
private MergeSort(){}
public static<E extends Comparable<E>> void sortOptimised1(E[] arr){
sortOptimised1(arr, 0, arr.length - 1);
}
private static<E extends Comparable<E>> void sortOptimised1(E[] arr, int left, int right){
if (left >= right){
return;
}
int mid = left + (right - left) / 2;
sortOptimised1(arr, left, mid);
sortOptimised1(arr, mid + 1, right);
/**
* 优化一:对有序数组排序的优化,合并前先判断一下是否需要合并
*/
if (arr[mid].compareTo(arr[mid + 1]) > 0) {
merge(arr, left, mid, right);
}
}
public static<E extends Comparable<E>> void merge(E[] arr, int left, int mid, int right) {
int i = left;
int j = mid + 1;
E[] tem = Arrays.copyOfRange(arr, left, right + 1);
for (int n = left; n < right + 1; n++) {
if (i == mid + 1){
arr[n] = tem[j - left];
j++;
}
else if (j == right + 1) {
arr[n] = tem[i - left];
i++;
}
else if (tem[i - left].compareTo(tem[j - left]) <= 0) {
arr[n] = tem[i - left];
i++;
}
else{
arr[n] = tem[j - left];
j++;
}
}
}
}
优化二:使用插入排序法优化(不稳定)
虽然归并排序法的时间复杂度小于插入排序法,但是merge()方法包含大量的if判断和赋值语句,常数级别的语句很多,在子区间长度较小时,可用插入排序法来代替
import java.util.Arrays;
public class Algorithm {
public static void main(String[] args) {
Integer[] arr = {6,6,3,2,4,7};
MergeSort.sortOptimised2(arr);
System.out.println(Arrays.toString(arr));
}
}
class MergeSort {
private MergeSort(){}
/**
* 优化二:在子区间长度小于16时,使用插入排序法代替
*/
public static<E extends Comparable<E>> void sortOptimised2(E[] arr){
sortOptimised2(arr, 0, arr.length - 1);
}
private static<E extends Comparable<E>> void sortOptimised2(E[] arr, int left, int right){
if (right - left <= 15) {
InsertionSort.sort(arr, left, right);
return;
}
int mid = left + (right - left) / 2;
sortOptimised2(arr, left, mid);
sortOptimised2(arr, mid + 1, right);
if (arr[mid].compareTo(arr[mid + 1]) > 0) {
merge(arr, left, mid, right);
}
}
public static<E extends Comparable<E>> void merge(E[] arr, int left, int mid, int right) {
int i = left;
int j = mid + 1;
E[] tem = Arrays.copyOfRange(arr, left, right + 1);
for (int n = left; n < right + 1; n++) {
if (i == mid + 1){
arr[n] = tem[j - left];
j++;
}
else if (j == right + 1) {
arr[n] = tem[i - left];
i++;
}
else if (tem[i - left].compareTo(tem[j - left]) <= 0) {
arr[n] = tem[i - left];
i++;
}
else{
arr[n] = tem[j - left];
j++;
}
}
}
}
class InsertionSort {
private InsertionSort() {}
/**
* 插入排序法修改一下传入的参数
*/
public static <E extends Comparable> void sort(E[] arr, int left, int right) {
for (int i = left + 1; i < right + 1; i++) {
E tem = arr[i];
int j;
for (j = i; j > left && tem.compareTo(arr[j - 1]) < 0; j--) {
arr[j] = arr[j - 1];
}
arr[j] = tem;
}
}
}
优化三:内存优化
merge()方法在每次调用时,都会开辟一个新的数组来接收传来的数组,当数据规模很大时,空间消耗也会损失很多性能
因此在排序之前先创建一个原数组的副本temp,merge()方法就在temp中进行赋值和读取,节省了大量的空间
import java.util.Arrays;
public class Algorithm {
public static void main(String[] args) {
Integer[] arr = {6,6,3,2,4,7};
MergeSort.sortOptimised3(arr);
System.out.println(Arrays.toString(arr));
}
}
class MergeSort {
private MergeSort(){}
/**
* 优化三:内存操作优化,提前将arr数组保存一个副本,当作参数传入,不用每次再新建临时数组,也没有索引的偏差了
*/
public static<E extends Comparable<E>> void sortOptimised3(E[] arr){
E[] temp = Arrays.copyOf(arr, arr.length);
sortOptimised3(arr, 0, arr.length - 1, temp);
}
private static<E extends Comparable<E>> void sortOptimised3(E[] arr, int left, int right, E[] temp){
if (left >= right){
return;
}
int mid = left + (right - left) / 2;
sortOptimised3(arr, left, mid, temp);
sortOptimised3(arr, mid + 1, right, temp);
if (arr[mid].compareTo(arr[mid + 1]) > 0) {
mergeOptimised(arr, left, mid, right, temp);
}
}
public static<E extends Comparable<E>> void mergeOptimised(E[] arr, int left, int mid, int right, E[] temp) {
int i = left;
int j = mid + 1;
/**
* System.arraycopy()方法将传过来的有序子区间在相同位置赋值给副本数组temp,因此索引范围一致没有偏移
*/
System.arraycopy(arr, left, temp, left, right - left + 1);
for (int n = left; n < right + 1; n++) {
if (i == mid + 1){
arr[n] = temp[j];
j++;
}
else if (j == right + 1) {
arr[n] = temp[i];
i++;
}
else if (temp[i].compareTo(temp[j]) <= 0) {
arr[n] = temp[i];
i++;
}
else{
arr[n] = temp[j];
j++;
}
}
}
}