Python implementation of Mergesort for Linked list doesn't work

问题

I couldn't find a simple implementation of Merge Sort in Python for Linked Lists anywhere. Here's what I tried:

Definition for singly-linked list:

class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None 

Merge Sort Implementation:

def mergeSortLinkedList(A):
    # Base case length of 0 or 1:
    if A == None or A.next == None:
        return A

    leftHalf, rightHalf = splitTheList(A)

    mergeSortLinkedList(leftHalf)
    mergeSortLinkedList(rightHalf)

    # The above two lines should be modified to the following. Thanks to the answers.

    # leftHalf  = mergeSortLinkedList(leftHalf)
    # rightHalf = mergeSortLinkedList(rightHalf)

    return mergeTheLists(leftHalf, rightHalf)

Split:

def splitTheList(sourceList):
    if sourceList == None or sourceList.next == None:
        leftHalf = sourceList
        rightHalf = None

        return leftHalf, rightHalf

    else:
        midPointer = sourceList
        frontRunner = sourceList.next
        # totalLength += 1        - This is unnecessary

        while frontRunner != None:
            frontRunner = frontRunner.next

            if frontRunner != None:
                frontRunner = frontRunner.next
                midPointer = midPointer.next

    leftHalf = sourceList
    rightHalf = midPointer.next
    midPointer.next = None

    return leftHalf, rightHalf

Merge:

def mergeTheLists(leftHalf, rightHalf):
    fake_head = ListNode(None)
    curr = fake_head

    while leftHalf and rightHalf:
        if leftHalf.val < rightHalf.val:
            curr.next = leftHalf
            leftHalf = leftHalf.next

        else:
            curr.next = rightHalf
            rightHalf = rightHalf.next

        curr = curr.next

    if leftHalf == None:
        curr.next = rightHalf

    elif rightHalf == None:
        curr.next = leftHalf

    return fake_head.next

Data:

# Node A:
nodeA1 = ListNode(2)

nodeA2 = ListNode(1)
nodeA1.next = nodeA2

nodeA3 = ListNode(9)
nodeA2.next = nodeA3

nodeA4 = ListNode(3)
nodeA3.next = nodeA4

# Node C:
nodeC1 = ListNode(5)
nodeA4.next = nodeC1

nodeC2 = ListNode(6)
nodeC1.next = nodeC2

nodeC3 = ListNode(4)
nodeC2.next = nodeC3

nodeC4 = ListNode(5)
nodeC3.next = nodeC4

Expected output when mergeSortLinkedList(nodeA1) is called:

1 2 3 4 5 5 6 9

I get the following instead:

2 5 6 9

I am unable to figure out where the miss is. Please help.

回答1:

You don't use the return values from the recursive call. The code should be:

def mergeSortLinkedList(A):
    if A is None or A.next is None:
        return A

    leftHalf, rightHalf = splitTheList(A)

    left = mergeSortLinkedList(leftHalf)
    right = mergeSortLinkedList(rightHalf)

    return mergeTheLists(left, right)

After the call of the function, the argument in some cases doesn't point to the head of the sorted list.

The next bug in the code is usage of undefined variable totalLength.

回答2:

Try changing the lines in the mergeSort function to:

leftHalf = mergeSortLinkedList(leftHalf)
rightHalf = mergeSortLinkedList(rightHalf)

来源：https://stackoverflow.com/questions/39337414/python-implementation-of-mergesort-for-linked-list-doesnt-work