在 Python 中实现归并排序

Question

I'm trying to implement simple merge sort algorithm in Python.

arr=[1,3,5,2,4,6]
n=6
l=0
h=n-1

def merge_Sort(l,h):
    if(l==h):
        return arr[l]

    m=(h+l)//2
    arr1=merge_Sort(l,m)
    arr2=merge_Sort(m+1,h)

    s1=m-l
    s2=h-(m+1)
    mer=[]
    k1=k2=0

    while(k1<=s1 or k2<=s2):
        if(arr1[k1] < arr2[k2]):
            mer.append(arr1[k1])
            k1+=1
        else:
            mer.append(arr2[k2])
            k2+=1

    if(k1>s1):
        while(k2<=s2):
            mer.append(arr2[k2])
            k2+=1

    if(k2>s2):
        while(k1<=s1):
            mer.append(arr1[k1])
            k1+=1    

    return mer

res=merge_Sort(l,h)

print(res)

But I'm getting this error message when running the above code:

TypeError: 'int' object is not subscriptable

Can anybody explain to me why I'm geting this error?

Answer 1

The problems are here:

    if(l==h):
        return arr[l]
    ...
    while(k1<=s1 or k2<=s2):

Use this code instead:

    if(l==h):
        return arr[l:l+1]
    ...
    while(k1<=s1 and k2<=s2):

---

Side note, this piece of code:

    if(k1>s1):
        while(k2<=s2):
            mer.append(arr2[k2])
            k2+=1

    if(k2>s2):
        while(k1<=s1):
            mer.append(arr1[k1])
            k1+=1

can be simplified to this:

    mer.extend(arr1[k1:s1+1])
    mer.extend(arr2[k2:s2+1])

Answer 2

Change this:

    if(l==h):
        return arr[l]

into:

    if(l==h):
        return [arr[l]]

and this:

    while(k1<=s1 or k2<=s2):

into:

    while(k1<=s1 and k2<=s2):

I tested it and it works fine.

Answer 3

TypeError: 'int' object is not subscriptable means you're trying to iterate (or use something like a[0] ) over int .

Add this before the "crash line".

print(arr1, arr2)

Output: 1 3

So you're doing 3[k2] .

Answer 4

an alternative method would be :

def merge_sort(a):

if len(a) <= 1:
    return a
mid = len(a) // 2
left = merge_sort(a[:mid])
right = merge_sort(a[mid:])
return merge(left, right)

def merge(left,right):

if not left:
    return right
if not right:
    return left
if left[0] < right[0]:
    return [left[0]] + merge(left[1:], right)
return [right[0]] + merge(left, right[1:])

Answer 5

This would be a relatively standard method:

def merge(left, right):
    res = []

    while len(left) > 0 and len(right) > 0:
        if left[0] <= right[0]:
            res.append(left[0])
            left.pop(0)
        else:
            res.append(right[0])
            right.pop(0)

    while len(left) > 0:
        res.append(left[0])
        left.pop(0)
    while len(right) > 0:
        res.append(right[0])
        right.pop(0)

    return res


def merge_sort(lst):
    if len(lst) <= 1:
        return lst

    left = lst[:(len(lst)//2)]
    right = lst[(len(lst)//2):]

    left = merge_sort(left)
    right = merge_sort(right)

    return merge(left,right)