Python 3 中的递归

Question

我试图以不同的方式进行合并排序（而不是文本中可用的方式等等），现在我有一个成功的合并算法，但问题是当我递归调用半列表时，合并的东西没有得到更新。帮我解决递归中的可变生命问题。

下面给出的代码的输出是：-

[9, 5]
After merging [5, 9]
[12, 4]
After merging [4, 12]
[9, 5, 12, 4] #it should be (the updated one i.e  [5, 9, 4, 12])
After merging [5, 9, 12, 4]
[6, 8]
After merging [6, 8]
[45, 2]
After merging [2, 45]
[6, 8, 45, 2]
After merging [6, 8, 45, 2]
[9, 5, 12, 4, 6, 8, 45, 2]
After merging [4, 5, 6, 8, 9, 12, 45, 2]
[9, 5, 12, 4, 6, 8, 45, 2]

def merge(arr1, arr2):
"""
Input is two sorted lists
Output is a single merged list
"""
        for i in arr1:
            for j in list(range(len(arr2))):
                if i<arr2[j]:
                    arr2.append(arr2[-1])
                    for count in list(range(len(arr2)-1, j, -1)):
                        arr2[count] = arr2[count-1]
                    arr2[j] = i
                    break
                if j == len(arr2)-1:
                    arr2.append(i)
        return arr2
    def mergeSort(arr):
        if len(arr) !=1:
            mergeSort(arr[:len(arr)//2])
            mergeSort(arr[len(arr)//2:])
            print(arr)
            arr = merge(arr[:len(arr)//2],arr[len(arr)//2:])
            print("After merging", arr)

        else:
            return arr
    a = [9,5,12, 4, 6, 8,45, 2]
    mergeSort(a)
    print(a)

Answer 1

您应该将 mergeSort return 合并为列表，调用者应该输出 mergeSort 的 returning 值：

def merge(arr1, arr2):
    merged = []
    while arr1 and arr2:
        if arr1[0] > arr2[0]:
            arr1, arr2 = arr2, arr1
        merged.append(arr1.pop(0))
    merged.extend(arr1 or arr2)
    return merged
def mergeSort(arr):
    if len(arr) <= 1:
        return arr
    return merge(mergeSort(arr[:len(arr)//2]), mergeSort(arr[len(arr)//2:]))
a = [9, 5, 12, 4, 6, 8, 45, 2]
print(mergeSort(a))

Answer 2

我只做了很少的改动，所以代码仍然是你的，你是如此接近，看：

问题是，在检查你的代码之后，你必须将两半传递给 merge 方法，还要检查它是否为空。此外，return 的结果要好得多，而不是在原地进行更改-

def merge(arr1, arr2):
        for i in arr1:
            for j in list(range(len(arr2))):
                if i<arr2[j]:
                    arr2.append(arr2[-1])
                    for count in list(range(len(arr2)-1, j, -1)):
                        arr2[count] = arr2[count-1]
                    arr2[j] = i
                    break
                if j == len(arr2)-1:
                    arr2.append(i)
        return arr2
def mergeSort(arr):
    if len(arr) !=1 and len(arr):
        ary1 = mergeSort(arr[:len(arr)//2])
        ary2 = mergeSort(arr[len(arr)//2:])
        print(arr)
        ary3 = merge(ary1,ary2)
        print("After merging", ary3)
        return ary3
    else:
        return arr
a = [9,5,12, 4, 6, 8,45, 2]

print(mergeSort(a))

输出

[9, 5]
After merging [5, 9]
[12, 4]
After merging [4, 12]
[9, 5, 12, 4]
After merging [4, 5, 9, 12]
[6, 8]
After merging [6, 8]
[45, 2]
After merging [2, 45]
[6, 8, 45, 2]
After merging [2, 6, 8, 45]
[9, 5, 12, 4, 6, 8, 45, 2]
After merging [2, 4, 5, 6, 8, 9, 12, 45]
[2, 4, 5, 6, 8, 9, 12, 45]

Python 3 中的递归

Recursion in Python 3

python

recursion

mergesort

python-3.x