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算法原理

升序实现，反之则亦然：
1.重复遍历要排列的数列，比较相邻的元素。如果第一个比第二个大，交换
2.第一轮结束确定一个最大元素到末尾
3.针对除了最后一个元素外的所有元素重复上面步骤

python代码实现

arr = [86, 39, 77, 23, 32, 45, 58, 63, 93, 4, 37, 22]  # 待排序数组# 升序实现
for i in range(len(arr)):print("开始第{}轮排序：".format(i))for j in range(i, len(arr) - 1 - i):if arr[j + 1] < arr[j]:tmp = arr[j]arr[j] = arr[j+1]arr[j+1] = tmpprint(arr)  # 打印查看每一步排序后结果print("第{}排序轮后:".format(i))print(arr)

执行结果：
开始第0轮排序：
[39, 86, 77, 23, 32, 45, 58, 63, 93, 4, 37, 22]
[39, 77, 86, 23, 32, 45, 58, 63, 93, 4, 37, 22]
[39, 77, 23, 86, 32, 45, 58, 63, 93, 4, 37, 22]
[39, 77, 23, 32, 86, 45, 58, 63, 93, 4, 37, 22]
[39, 77, 23, 32, 45, 86, 58, 63, 93, 4, 37, 22]
[39, 77, 23, 32, 45, 58, 86, 63, 93, 4, 37, 22]
[39, 77, 23, 32, 45, 58, 63, 86, 93, 4, 37, 22]
[39, 77, 23, 32, 45, 58, 63, 86, 93, 4, 37, 22]
[39, 77, 23, 32, 45, 58, 63, 86, 4, 93, 37, 22]
[39, 77, 23, 32, 45, 58, 63, 86, 4, 37, 93, 22]
[39, 77, 23, 32, 45, 58, 63, 86, 4, 37, 22, 93]
第0排序轮后:
[39, 77, 23, 32, 45, 58, 63, 86, 4, 37, 22, 93]
开始第1轮排序：
[39, 23, 77, 32, 45, 58, 63, 86, 4, 37, 22, 93]
[39, 23, 32, 77, 45, 58, 63, 86, 4, 37, 22, 93]
[39, 23, 32, 45, 77, 58, 63, 86, 4, 37, 22, 93]
[39, 23, 32, 45, 58, 77, 63, 86, 4, 37, 22, 93]
[39, 23, 32, 45, 58, 63, 77, 86, 4, 37, 22, 93]
[39, 23, 32, 45, 58, 63, 77, 86, 4, 37, 22, 93]
[39, 23, 32, 45, 58, 63, 77, 4, 86, 37, 22, 93]
[39, 23, 32, 45, 58, 63, 77, 4, 37, 86, 22, 93]
[39, 23, 32, 45, 58, 63, 77, 4, 37, 22, 86, 93]
第1排序轮后:
[39, 23, 32, 45, 58, 63, 77, 4, 37, 22, 86, 93]
开始第2轮排序：
[39, 23, 32, 45, 58, 63, 77, 4, 37, 22, 86, 93]
[39, 23, 32, 45, 58, 63, 77, 4, 37, 22, 86, 93]
[39, 23, 32, 45, 58, 63, 77, 4, 37, 22, 86, 93]
[39, 23, 32, 45, 58, 63, 77, 4, 37, 22, 86, 93]
[39, 23, 32, 45, 58, 63, 4, 77, 37, 22, 86, 93]
[39, 23, 32, 45, 58, 63, 4, 37, 77, 22, 86, 93]
[39, 23, 32, 45, 58, 63, 4, 37, 22, 77, 86, 93]
第2排序轮后:
[39, 23, 32, 45, 58, 63, 4, 37, 22, 77, 86, 93]
开始第3轮排序：
[39, 23, 32, 45, 58, 63, 4, 37, 22, 77, 86, 93]
[39, 23, 32, 45, 58, 63, 4, 37, 22, 77, 86, 93]
[39, 23, 32, 45, 58, 4, 63, 37, 22, 77, 86, 93]
[39, 23, 32, 45, 58, 4, 37, 63, 22, 77, 86, 93]
[39, 23, 32, 45, 58, 4, 37, 22, 63, 77, 86, 93]
第3排序轮后:
[39, 23, 32, 45, 58, 4, 37, 22, 63, 77, 86, 93]
开始第4轮排序：
[39, 23, 32, 45, 4, 58, 37, 22, 63, 77, 86, 93]
[39, 23, 32, 45, 4, 37, 58, 22, 63, 77, 86, 93]
[39, 23, 32, 45, 4, 37, 22, 58, 63, 77, 86, 93]
第4排序轮后:
[39, 23, 32, 45, 4, 37, 22, 58, 63, 77, 86, 93]
开始第5轮排序：
[39, 23, 32, 45, 4, 22, 37, 58, 63, 77, 86, 93]
第5排序轮后:
[39, 23, 32, 45, 4, 22, 37, 58, 63, 77, 86, 93]
开始第6轮排序：
第6排序轮后:
[39, 23, 32, 45, 4, 22, 37, 58, 63, 77, 86, 93]
开始第7轮排序：
第7排序轮后:
[39, 23, 32, 45, 4, 22, 37, 58, 63, 77, 86, 93]
开始第8轮排序：
第8排序轮后:
[39, 23, 32, 45, 4, 22, 37, 58, 63, 77, 86, 93]
开始第9轮排序：
第9排序轮后:
[39, 23, 32, 45, 4, 22, 37, 58, 63, 77, 86, 93]
开始第10轮排序：
第10排序轮后:
[39, 23, 32, 45, 4, 22, 37, 58, 63, 77, 86, 93]
开始第11轮排序：
第11排序轮后:
[39, 23, 32, 45, 4, 22, 37, 58, 63, 77, 86, 93]

Java实现

public class BubbleSort {public static void bubbleSort(int [] arr) {int length = arr.length;for (int i = 0; i < length - 1; i++) {for (int j = i; j < length - 1; j++) {if (arr[j + 1] < arr[j]) {int tmp = arr[j];arr[j] = arr[j + 1];arr[j + 1] = tmp;}}}}public static void main(String[] args) {int[] arr = new int[]{1, 8, 2, 5, 4, 9, 10, 6};bubbleSort(arr);for (int a :arr) {System.out.println(a);}}
}

总结

每一轮会确定一个元素的最终位置，时间复杂度为O(n^2)

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