如何在python中一次性找到链表的中间元素？

Question

I am trying to solve a linked list problem, to find the middle element in a single pass using python. Could someone please review my code and suggest the best manner to do this?

  class Node(object):
      def __init__(self, data=None, next=None):
          self.data = data
          self.next = next
      def __str__(self):
          return str(self.data)

  def print_nodes(node):
      while node:
          print node
          node = node.next

  def find_middle(node):
      while node:
          current = node
          node = node.next
          second_pointer = node.next
          next_pointer = second_pointer.next
          if next_pointer is None:
              return "Middle node is %s" % str(current)

  node1 = Node(1)
  node2 = Node(2)
  node3 = Node(3)
  node4 = Node(4)
  node5 = Node(5)

  node1.next = node2
  node2.next = node3
  node3.next = node4
  node4.next = node5

  print find_middle(node1)

Answer 1

I merged all the methods for you creating, finding and printing.

class Node(object):
    def __init__(self, data=None, next=None):
        self.data = data
        self.next = next
    def __str__(self):
        return str(self.data)

def create_linked_list(n):
    """Creating linked list for the given
       size"""
    linked_list = Node(1)
    head = linked_list
    for i in range(2, n):
        head.next = Node(i)
        head = head.next
    return linked_list

def print_linked_list(node):
    """To print the linked list in forward"""
    while node:
        print '[',node,']','[ref] ->',
        node = node.next
    print '-> None'

def find_middle1(node):
    tick = False
    half = node
    while node:
        node = node.next
        if tick:
            half = half.next
        tick = not tick
    return "Middle node is %s" % str(half)

def find_middle2(node):
    list = []
    while node:
        list.append(node)
        node = node.next
    return "Middle node is %s" % str(list[len(list)/2])


node = create_linked_list(10)
print_linked_list(node)

print find_middle1(node)
print find_middle2(node)

Output:

[ 1 ] [ref] -> [ 2 ] [ref] -> [ 3 ] [ref] -> [ 4 ] [ref] -> [ 5 ] [ref] -> [ 6 ] [ref] -> [ 7 ] [ref] -> [ 8 ] [ref] -> [ 9 ] [ref] -> -> None
Middle node is 5
Middle node is 5

Answer 2

Here's on way, it's one pass, though probably not as efficient as you'd like:

def find_middle(node):
    list = []
    while node:
        list.append(node)
        node = node.next
    return list[len(list)/2]

does that work?

Answer 3

You could keep two pointers, one that moves half as fast as the other.

def find_middle(node):
    tick = False
    half = node
    while node:
        node = node.next
        if (tick):
            half = half.next
        tick = not tick
    return "Middle node is %s" % str(half)

Answer 4

pseudo code for finding middle element of linked list : -

fast = head
slow = head


while(fast!=null) {



 if(fast.next!=null) {

      fast = fast.next.next
      slow = slow.next
 }

 else { 

  break
 }
}

// middle element
return slow

Answer 5

All of the above answers are right but for me, this worked best:

        def middleNode(self, head: ListNode) -> ListNode:
            list=[]
            while head:
                list.append(head)
                head=head.next
            return list[floor(len(list)/2)]

Here, using floor helped me otherwise my code gave me errors.

Answer 6

OK, this is NOT a good idea. But it DOES satisfy the constraint of traversing only once. Instead of traversing once (and a secondary half traversal), it (ab)uses the stack to emulate the half traversal (backwards instead of forwards). It's not a good idea, because Python doesn't have an infinitely growable stack (I wish Python would take a cue from the Smalltalk guys on this one), so you really can only handle lists in the size of hundreds, definitely not thousands (this is Python3, btw).

First I modified your script to build bigger lists with the change of a value:

last = root = Node(1)
for i in range(2, 312):
    node = Node(i)
    last.next = node
    last = node

Since we're using the stack and recursion, we need a way to return abruptly out of a deep call stack. So we create an Exception subclass, which is really more of a "notification" than an "exception".

class FoundMiddleNode(Exception):
    def __init__(self, node):
        super().__init__()
        self.node = node

Now for the recursive function:

def probeForMiddle(node, length):
    if node.next is None: #recursion stopper
        return length
    #call recursively
    lengthToEnd = probeForMiddle(node.next, length + 1)
    #is my distance from start half of what is being returned recursively as the full length?
    if (lengthToEnd // 2) - length == 0: 
        raise FoundMiddleNode(node) #throw an exception to abort the rest of the stack walk
    return lengthToEnd

And to use it we do:

try:
    probeForMiddle(root, 0)
except FoundMiddleNode as e:
    print(e.node)

Not pretty. Not a good idea in anything approximating production code. But a decent example of (ab)using recursion and exceptions to fill the requirement of only traversing once.

Answer 7

The best way to find the middle node is to have two pointers:

 P1 = root
    P2 = root
    While not p2.next == Null:
    P1 =p1.next
    P2 =P2.next.next

Answer 8

//Linked list

ll = {'A': ["data", "B"],
 'B': ["data", "C"],
 'C': ["data", "D"],
 'D': ["data", "E"],
 'E': ["data", None]}


def find_next_node(node="A"):
    return ll[node][1] if ll[node][1] else None

def find_mid_node(head="A"):
    slow = head
    fast = head
    while(fast!=None):
        for i in range(2):
            if find_next_node(fast):
                fast = find_next_node(node=fast)
            else:
                return slow

        for j in range(1):
            slow = find_next_node(node=slow)


print (find_mid_node())

Answer 9

You can write a smaller code for finding the middle node. Showing the snippet below:

   def find_middle(node):
      half = node
      while node and node.next is not None:
         node = node.next.next
         half = half.next
      return half

Few important points:

1. Logic will remain the same of keeping two pointers, one fast and other slow.
2. Write a Linked list class to create a linked list with the help of loop rather than explicitly setting next pointers.

Answer 10

This is very similar to what James and Jordan have posted already, it's just little simpler in terms of what it does and I've added explanation as comments to what's actually doing

class Node:
  def __init__(self, data=None, next=None):
    self.data = data
    self.next = next 
# loop through the items and check for next items using two pointers (fast and slow)
# set the speed for fast twice higher than the slow pointer so when when fast reaches
# the end the slow would be in the middle
def find_middle(head ):
    fast = slow = head 
    #slow = head
    while fast.next != None and fast.next.next != None:
        fast = fast.next.next
        slow = slow.next

    # slow is now at the middle :)
    print (slow.data  )


#setup data
node1 = Node(1)
node2 = Node(2)
node3 = Node(3)
node4 = Node(4)
node5 = Node(5)

node1.next = node2
node2.next = node3
node3.next = node4
node4.next = node5

find_middle(node1)

Answer 11

I might be little late but this works best for me. Writing complete code for creating, finding the midpoint & printing the linked list.

class Node:
    '''
    Class to create Node and next part of the linked list
    '''
    def __init__(self,data):
        self.data = data
        self.next = None
        

def createLL(arr):
    '''
    linked list creation
    '''
    head = Node(arr[0])
    tail = head
    for data in arr[1:]:
        tail.next = Node(data)
        tail = tail.next
        
    return head

def midPointOfLL(head):
    '''
    Here, i am using two pointers slow and fast, So idea is fast will be 2 times faster than slow pointer
    and when fast reaches the end of the linked list slow pointer will be exactly in the middle.
    '''

    slow = head
    fast = head
    if head is not None:
        while (fast.next is not None) and (fast.next.next is not None):
            slow = slow.next
            fast = fast.next.next
        
    return slow

def printLL(head):
    curr = head
    while curr :
        print(curr.data,end = "-->")
        curr = curr.next
    print('None')


arr  = list(map(int,input().split())) 
head = createLL(arr)       
midPoint = midPointOfLL(head)
print(midPoint.data)
printLL(head)

Answer 12

list=['ok','go','no','poi','duo','ok','go','nah']
b=0
b=int(len(list)/2) #print the middle element change .5 values 
print(list[b])
if((len(list)%2)==0): #if list is of even number print both middle values
    print(list[b+1])

when i searched this question i was looking for something like this so i can get both middle values when entered even num of elements

there must be a better way to do it i just start coding from like 3,4 day