【经典排序算法 time: 2023-10-12】冒泡排序（层层优化改进）

原理
每次循环找出一个最大的元素（动态演示）
第一版冒泡

java 复制代码

public class Maopao1 {
    public static void main(String[] args) {
        long start= System.currentTimeMillis();
        int[] arr2 = {11, 23, 69, 99, 1, 3, 45, 67, 5, 234, 678, 999, 7, 123};
        int[] result = maopao(arr2);
        System.out.println(Arrays.toString(result));

        System.out.println(System.currentTimeMillis()-start);
    }

    private static int[] maopao(int[] arr2) {
        for (int i = 0; i < arr2.length - 1; i++) {
            for (int j = 0; j < arr2.length - 1; j++) {
                if (arr2[j] > arr2[j + 1]) {
                    swap(arr2, j, j + 1);
                }
            }
            System.out.println(Arrays.toString(arr2));
        }
        System.out.println("循环了" +( arr2.length-1) + "次");
        return arr2;
    }

    public static void swap(int[] a, int i, int j) {
        int t = a[i];
        a[i] = a[j];
        a[j] = t;
    }
}

改进一：内层循环的次数与外层循环有关系，将a.length-1变成a.length-1-i，可以每次循环后减少一次循环次数。

java 复制代码

public class Maopao2 {
    public static void main(String[] args) {

        long start = System.currentTimeMillis();
        int[] arr2 = {11, 23, 69, 99, 1, 3, 45, 67, 5, 234, 678, 999, 7, 123};
        int[] result = maopao(arr2);
        System.out.println(Arrays.toString(result));
        System.out.println(System.currentTimeMillis()-start);
    }

    private static int[] maopao(int[] arr2) {
        for (int i = 0; i < arr2.length - 1; i++) {
            for (int j = 0; j < arr2.length - 1 - i; j++) {
                if (arr2[j] > arr2[j + 1]) {
                    swap(arr2, j, j + 1);
                }
            }
            System.out.println(Arrays.toString(arr2));
        }
        return arr2;
    }

    public static void swap(int[] a, int i, int j) {
        int t = a[i];
        a[i] = a[j];
        a[j] = t;
    }
}

改进二，只要一轮比较没有发生交换，则可以提前结束冒泡，在本身有序的序列情况下，循环一次就行了

java 复制代码

public class Maopao2 {
    public static void main(String[] args) {

        long start = System.currentTimeMillis();
        int[] arr2 = {11, 23, 69, 99, 1, 3, 45, 67, 5, 234, 678, 999, 7, 123};
        int[] result = maopao(arr2);
        System.out.println(Arrays.toString(result));
        System.out.println(System.currentTimeMillis()-start);
    }

    private static int[] maopao(int[] arr2) {
        for (int i = 0; i < arr2.length - 1; i++) {
            boolean swapBool = false; // 发生交换
            for (int j = 0; j < arr2.length - 1 - i; j++) {
                if (arr2[j] > arr2[j + 1]) {
                    swap(arr2, j, j + 1);
                    swapBool = true;
                }
            }
            if (!swapBool) {
                System.out.println("循环了" + i + "次");
                break;
            }
            System.out.println(Arrays.toString(arr2));
        }
        return arr2;
    }

    public static void swap(int[] a, int i, int j) {
        int t = a[i];
        a[i] = a[j];
        a[j] = t;
    }
}

改进三，只需要内层循环，外层使用while（ture），用内层最后一个交换的索引来退出循环，如果最后交换的索引等于0的话说明没有交换，推出循环；

java 复制代码

public class Maopao3 {
    public static void main(String[] args) {

        long start = System.currentTimeMillis();
        int[] arr2 = {11, 23, 69, 99, 1, 3, 45, 67, 5, 234, 678, 7, 123,999};
        int[] result = maopao(arr2);
        System.out.println(Arrays.toString(result));

        System.out.println(System.currentTimeMillis()-start);
    }

    private static int[] maopao(int[] arr2) {
        int n = arr2.length - 1;
        int num = 0;
        while (true) {
            int last = 0; // 最后一次交换的索引
            for (int j = 0; j < n; j++) {
                if (arr2[j] > arr2[j + 1]) {
                    swap(arr2, j, j + 1);
                    last = j;
                }
            }
            n = last;
            num++;
            if (n == 0) {
                break;
            }
            System.out.println(Arrays.toString(arr2));
        }
        System.out.println("循环了" + num + "次");
        return arr2;
    }

    public static void swap(int[] a, int i, int j) {
        int t = a[i];
        a[i] = a[j];
        a[j] = t;
    }
}