归并排序法01:实现归并排序法

归并排序法是一种高级排序算法

核心思想:归并排序法是一种分治算法,将一个数组递归的一分为二,二分为四...定义一个merge()方法对两个子数组分别进行遍历比较,选出较小的元素放在前面,最终完成排序

递归实现归并排序法

import java.util.Arrays;

public class Algorithm {

    public static void main(String[] args) {

        Integer[] arr = {6,6,3,2,4,7};

        MergeSort.sort(arr);
        System.out.println(Arrays.toString(arr));
    }
}

class MergeSort {

    private MergeSort(){}

    /**
     * 用户只需要传入一个数组即可
     */
    public static<E extends Comparable<E>> void sort(E[] arr){
        
        sort(arr, 0, arr.length - 1);
    }

    /**
     * 为了实现递归,必须要有拆分数组的边界索引,因此定义一个私有、对用户不可见的方法来进行递归
     */
    private static<E extends Comparable<E>> void sort(E[] arr, int left, int right){

        /**
         * 如果元素个数小于等于1,不需要进行操作
         */
        if (left >= right){
            
            return;
        }

        /**
         * left + right有可能会产生整型溢出,因此写成如下方式可以避免出现这种错误
         */
        int mid = left + (right - left) / 2;

        /**
         * 递归的拆分区间
         */
        sort(arr, left, mid);
        sort(arr, mid + 1, right);

        /**
         * 对有序的两个子区间进行最终的排序
         */
        merge(arr, left, mid, right);
    }

    /**
     * 有序子区间arr[left, mid]和arr[mid + 1, right]
     */
    public static<E extends Comparable<E>> void merge(E[] arr, int left, int mid, int right) {

        /**
         * left和mid + 1是两个子数组的开始索引,表示当前进行比较的元素索引
         */
        int i = left;
        int j = mid + 1;

        /**
         * 因为对arr数组原地排序,数组元素会动态发生变化,不能正常读取下一个元素。因此先保存一下两个有序的子区间副本,根据副本的顺序来读取元素
         * 副本只需要保存当前的子区间[left, right + 1]而不是整个arr数组,否则空间复杂度会很大
         * Arrays.copyOfRange()方法按指定范围拷贝数组,注意结束索引是不包含的
         */
        E[] tem = Arrays.copyOfRange(arr, left, right + 1);

        for (int n = left; n < right + 1; n++) {

            /**
             * 如果i和j走到了自己小区间的边界,说明这个子区间已经排好序了,剩下的就是将另一个子区间的元素直接添加进来
             * 因为tem数组的索引是从0开始的,而我们要操作的arr数组是从left开始的,tem[0] == arr[left],i和j是由arr数组赋值的,所以跟tem有一个left的偏差
             */
            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++;
            }
        }
    }
}

拓展:利用辅助方法,打印递归的分步信息

import java.util.Arrays;

public class Algorithm {

    public static void main(String[] args) {

        Integer[] arr = {6,6,3,5,4,7,4,5,6,7,9};

        MergeSort.sort(arr, 0);
        System.out.println(Arrays.toString(arr));
    }
}

class MergeSort {

    private MergeSort(){}

    public static<E extends Comparable<E>> void sort(E[] arr, int depth){

        sort(arr, 0, arr.length - 1, depth);
    }

    private static<E extends Comparable<E>> void sort(E[] arr, int left, int right, int depth){

        String depthScale = depth(depth);

        System.out.print(depthScale);
        System.out.println(String.format("当前要排序的数组是arr[%d, %d]", left, right));

        if (left >= right){

            return;
        }

        int mid = left + (right - left) / 2;

        sort(arr, left, mid, depth++);
        sort(arr, mid + 1, right, depth++);

        System.out.print(depthScale);
        System.out.println(String.format("当前排序完成的数组是arr[%d, %d]和arr[%d, %d]", left, mid, mid + 1, right));

        merge(arr, left, mid, right, depth++);

        System.out.print(depthScale);
        System.out.println(String.format("当前合并完成的数组是arr[%d, %d]", left, right));
    }

    public static<E extends Comparable<E>> void merge(E[] arr, int left, int mid, int right, int depth) {

        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++;
            }
        }
    }

    /**
     * 打印分隔符,递归深度越深,分隔符越长
     */
    public static String depth(int depth){

        StringBuilder str = new StringBuilder();

        for (int i = 0; i < depth; i++) {
            str.append("----");
        }

        return str.toString();
    }
}

复杂度分析

从归并排序的递归树来看,递归深度为log2n,而每层所要处理的元素都是n个

因此归并排序的时间复杂度为O(nlogn)

在这里插入图片描述

选择、插入、归并排序法性能比较

import java.util.Arrays;
import java.util.Random;

public class Algorithm {

    public static void main(String[] args) {

        Integer[] testScale = {10000, 100000};

        for (Integer n : testScale) {

            Integer[] randomArr = ArrayGenerator.generatorRandomArray(n, n);
            Integer[] sortedArr = ArrayGenerator.generatorSortedArray(n);

            Integer[] arr1 = Arrays.copyOf(randomArr, randomArr.length);
            Integer[] arr2 = Arrays.copyOf(randomArr, randomArr.length);

            Integer[] arr3 = Arrays.copyOf(sortedArr, sortedArr.length);
            Integer[] arr4 = Arrays.copyOf(sortedArr, sortedArr.length);

            System.out.println("测试随机数组排序性能");
            System.out.println();

            Verify.testTime("SelectionSort", randomArr);
            Verify.testTime("InsertionSort", arr1);
            Verify.testTime("MergeSort", arr2);

            System.out.println();

            System.out.println("测试有序数组排序性能");
            System.out.println();

            Verify.testTime("SelectionSort", sortedArr);
            Verify.testTime("InsertionSort", arr3);
            Verify.testTime("MergeSort", arr4);

            System.out.println();
        }
    }
}

class MergeSort {

    private MergeSort(){}

    public static<E extends Comparable<E>> void sort(E[] arr){

        sort(arr, 0, arr.length - 1);
    }

    private static<E extends Comparable<E>> void sort(E[] arr, int left, int right){

        if (left >= right){

            return;
        }

        int mid = left + (right - left) / 2;

        sort(arr, left, mid);
        sort(arr, mid + 1, right);

        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() {}

    public static <E extends Comparable> void sort(E[] arr) {

        for (int i = 1; i < arr.length; i++) {

            E tem = arr[i];
            int j;

            for (j = i; j > 0 && tem.compareTo(arr[j - 1]) < 0; j--) {
                arr[j] = arr[j - 1];
            }

            arr[j] = tem;
        }
    }
}

class SelectionSort {

    private SelectionSort(){}

    public static<E extends Comparable<E>> void sort(E[] arr){

        for (int i = 0; i < arr.length - 1; i++) {

            int minIndex = i;

            for (int j = i + 1; j < arr.length; j++) {

                if (arr[j].compareTo(arr[minIndex]) < 0) {
                    minIndex = j;
                }
            }

            if (i != minIndex) {

                E tem;
                tem = arr[minIndex];
                arr[minIndex] = arr[i];
                arr[i] = tem;
            }
        }
    }
}

class ArrayGenerator {

    private ArrayGenerator (){}

    public static Integer[] generatorRandomArray (Integer n, Integer maxBound){

        Integer[] arr = new Integer[n];
        Random random = new Random();

        for (int i = 0; i < n; i++) {
            arr[i] = random.nextInt(maxBound);
        }

        return arr;
    }

    public static Integer[] generatorSortedArray (Integer n){

        Integer[] arr = new Integer[n];

        for (int i = 0; i < n; i++) {
            arr[i] = i;
        }

        return arr;
    }
}

class Verify {

    private Verify (){}

    public static<E extends Comparable<E>> boolean isSorted(E[] arr){

        for (int i = 0; i < arr.length - 1; i++) {

            if (arr[i].compareTo(arr[i + 1]) > 0) {
                return false;
            }
        }

        return true;
    }

    public static<E extends Comparable<E>> void testTime(String AlgorithmName, E[] arr) {

        long startTime = System.nanoTime();

        if (AlgorithmName.equals("SelectionSort")){
            SelectionSort.sort(arr);
        }

        if (AlgorithmName.equals("InsertionSort")) {
            InsertionSort.sort(arr);
        }

        if (AlgorithmName.equals("MergeSort")) {
            MergeSort.sort(arr);
        }

        long endTime = System.nanoTime();

        if (!Verify.isSorted(arr)){
            throw new RuntimeException(AlgorithmName + "算法排序失败!");
        }

        System.out.println(String.format("%s算法,测试用例为%d,执行时间:%f秒", AlgorithmName, arr.length, (endTime - startTime) / 1000000000.0));
    }
}
posted @ 2021-10-19 19:55  振袖秋枫问红叶  阅读(68)  评论(0)    收藏  举报