堆排序

java代码实现

package csdn.dreamzuora.sort;import java.util.List;/*** Title: 抽象出排序类* Description:** @version 1.0* @author: weijie* @date: 2020/10/22 17:59*/
public abstract class Sort<E> {public void sort(List<E> array){if (array == null || array.isEmpty()){return;}};public void sort(List<E> array, int left, int right){};}

package csdn.dreamzuora.sort;import java.util.List;/*** Title: 堆排序* Description:** @version 1.0* @author: weijie* @date: 2020/10/22 17:51* @url https://www.cnblogs.com/rosesmall/p/9554545.html*/
public class HeapSort extends Sort<Integer>{@Overridepublic void sort(List<Integer> array) {//1.构建大顶堆for(int size = array.size(), i = size / 2 - 1; i >= 0; i--){//从第一个非叶子结点从下至上，从右至左调整结构adjustHeap(array, i, size);}//2.调整堆结构、交换堆顶元素与末尾元素for (int j = array.size() - 1; j > 0; j--){swap(array, 0, j);adjustHeap(array, 0, j);}}/*** 调整大顶堆（仅是调整过程，建立在大顶堆已构建的基础上）* @param arr* @param i* @param length*/public void adjustHeap(List<Integer> arr, int i, int length){//先取出当前元素iint temp = arr.get(i);//从i结点的左子结点开始，也就是2i+1处开始for(int k = i*2 + 1; k < length; k = k*2 + 1){//如果左子结点小于右子结点，k指向右子结点if(k+1 < length && arr.get(k) < arr.get(k+1)){k++;}//如果子节点大于父节点，将子节点值赋给父节点（不用进行交换）if(arr.get(k) > temp){arr.set(i, arr.get(k)) ;i = k;}else{break;}}//将temp值放到最终的位置arr.set(i, temp);}private void swap(List<Integer> arr, int a, int b){int temp = arr.get(a);arr.set(a, arr.get(b));arr.set(b, temp);}
}

单元测试

package csdn.dreamzuora.sort;import org.junit.Test;
import org.junit.jupiter.api.Assertions;import java.util.Arrays;
import java.util.List;import static org.junit.Assert.*;/*** Title:* Description:** @version 1.0* @author: weijie* @date: 2020/10/26 23:14*/
public class HeapSortTest {@Testpublic void sort() {HeapSort heapSort = new HeapSort();List<Integer> list = Arrays.asList(4, 5, 2, 6);heapSort.sort(list);Assertions.assertEquals(Arrays.asList(2, 4, 5, 6), list);}
}

归并排序

java代码实现

package csdn.dreamzuora.sort;import java.util.ArrayList;
import java.util.List;/*** Title: 归并排序* Description:* 将数组进行拆分，拆分后按照一定的顺序各自排序，然后再归并到一起，使得归并后的数组仍然有序** 归并排序算法可以利用递归的思想或者迭代的思想去实现。首先我们先把一个无序的数组去拆分，然后利用一定的规则，去合并。* 类似于二叉树的结构。其总的时间复杂度为O( n log n)。空间复杂度为 S（nlogn）** @version 1.0* @author: weijie* @date: 2020/10/22 17:51* @url: https://blog.csdn.net/dreamzuora/article/details/52830740*/
public class MergeSort extends Sort<Integer>{@Overridepublic void sort(List<Integer> a) {if (a == null || a.isEmpty()){return;}List<Integer> b = new ArrayList<>();for (int i = 0; i < a.size(); i++){b.add(0);}merge(a, b,0, a.size());for (int i = 0; i < b.size(); i++){a.set(i, b.get(i));}}private void merge(List<Integer> a, List<Integer> b, int start, int end){//递归出口if (end - start <= 1){return;}int mid = (start + end)/2;int left = start;int right = mid;int index = start;//递归入口merge(a, b, start, mid);merge(a, b, mid, end);//核心处理while (left < mid || right < end){//右区间遍历完成，直接取左区间值if (right >= end){b.set(index ++, a.get(left ++));}else if (left < mid && a.get(left) < a.get(right)){b.set(index ++, a.get(left ++));}else {b.set(index ++, a.get(right ++));}}//原数组赋值for (int i = start; i < index; i++){a.set(i, b.get(i));}}}

单元测试

package csdn.dreamzuora.sort;import org.junit.Test;
import org.junit.jupiter.api.Assertions;import java.util.*;import static org.junit.Assert.*;/*** Title:* Description:** @version 1.0* @author: weijie* @date: 2020/10/28 10:51*/
public class MergeSortTest {@Testpublic void sort() {MergeSort mergeSort = new MergeSort();Random random = new Random();List<Integer> actuallyList = Arrays.asList(random.nextInt(100), random.nextInt(100), random.nextInt(100), random.nextInt(100), random.nextInt(100), random.nextInt(100));List<Integer> expectList = new ArrayList<>(actuallyList);Collections.sort(expectList);mergeSort.sort(actuallyList);Assertions.assertEquals(expectList, actuallyList);}
}