Python3 中的递归冒泡排序 - 返回 "None"

Question

我在 Python 3 中创建了一个小函数来执行 BubbleSort（刚刚学习如何编码），但我找不到问题。

这是代码。它出于某种原因返回 "None"。有人可以看一下吗？谢谢！

arr = [1,5,2,7,3]

def bubbleSort(array):
    count = 0
    #print("array is currently",array)
    for idx in range(len(array)-1):
        if array[idx] > array[idx + 1]:
            array[idx],array[idx + 1] = array[idx + 1],array[idx]
            count += 1
            #print("swaped and count is currently",count)
            #print("array is currently",array)
    if count == 0:
        #print("Count is zero")
        #print("array is currently",array)
        return array
    else:
        #print("Count is not zero")
        bubbleSort(array)

print(bubbleSort(arr))

Answer 1

您需要return排序的数组

arr = [1,5,2,7,3]

def bubbleSort(array):
    count = 0
    #print("array is currently",array)
    for idx in range(len(array)-1):
        if array[idx] > array[idx + 1]:
            array[idx],array[idx + 1] = array[idx + 1],array[idx]
            count += 1
            #print("swaped and count is currently",count)
            #print("array is currently",array)
    if count == 0:
        #print("Count is zero")
        #print("array is currently",array)
        return array
    else:
        #print("Count is not zero")
        return bubbleSort(array)

print(bubbleSort(arr))

Answer 2

冒泡排序使用递归而不使用任何循环，

def bubble_sort_recur(a, i, j, n):
    if j == n:
        i = i+1
        j = 0
    if i == n:
        return 
    if a[i] > a[j]:
        temp = a[j]
        a[j] = a[i]
        a[i] = temp
        bubble_sort_recur(a, i, j+1, n);
    else:
        bubble_sort_recur(a, i, j + 1, n);
    return a

a = [1, 12, 3, 4]
a = bubble_sort_recur(a, 0, 0, len(a))
print(a)

Answer 3

代码行更少的最简单的解决方案：

def bubblesort(l,n):
    for i in range(len(l)-2):
        if l[i] > l[i+1]:
            l[i], l[i+1] = l[i+1], l[i]
    if n>1:        
        bubblesort(l,n-1)   

l = [6,2,9,11,9,3,7,12]
n=len(l)    
bubblesort(l,n)
print(l)

Answer 4

我真的很喜欢@user3065757 的方法！无论如何，我的代码是

def rec_bubble_sort(myList):
    flag = 0
    for i in range(len(myList)-1):
        if myList[i] > myList[i+1]:
            myList[i], myList[i+1] = myList[i+1], myList[i]
            flag = 1
    if flag:
        return rec_bubble_sort(myList[0:len(myList)-1])+myList[len(myList)-1:]
    else:
        return myList


myList = [2, 6, 4, 0, 5, 7, 2, 4, 1]
print(rec_bubble_sort(myList))

我的代码与@RohithS98 类似，但我使用过

return rec_bubble_sort(myList[0:len(myList)-1])+myList[len(myList)-1:]

原因是函数调用时不会遍历整个列表

Answer 5

public class RecursionBubbleSort {
    public static void main(String[] args) {
        int[] nums = {4, 3, 2, 1};
        int[] sorted = bubbleSort(nums, 1, nums.length);

        for(int i: sorted){
            System.out.print(i + " ");
        }
    }

    public static int[] bubbleSort(int[] nums, int start, int end){
        if(start >= end) return nums;

        nums = innerSort(nums, start, end);
        
        return bubbleSort(nums, start, end-1);
    }

    public static int[] innerSort(int[] nums, int start, int size){
        if(start >= size) return nums;

        if(nums[start-1] > nums[start]){
            int temp = nums[start];
            nums[start] = nums[start-1];
            nums[start-1] = temp;
        }

        return innerSort(nums, start+1, size);
    }
}

Python3 中的递归冒泡排序 - 返回 "None"

BubbleSort with recursion in Python3 - Returning "None"

python

bubble-sort

python-3.x