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【经典排序算法 time: 2023-10-12】冒泡排序（层层优化改进）

news 2025/7/18 18:43:37

原理
每次循环找出一个最大的元素（动态演示）
第一版冒泡

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。

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

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；

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;}
}