每日算法题 - Reverse a Linked List

Hi, here's your problem today. This problem was recently asked by Google:

Given a singly-linked list, reverse the list. This can be done iteratively or recursively. Can you get both solutions?

Example:
Input: 4 - 3 - 2 - 1 - 0 - NULL
Output: 0 - 1 - 2 - 3 - 4 - NULL

class ListNode(object):

  def __init__(self, x):

    self.val = x

    self.next = None

  

  # Function to print the list

  def printList(self):

    node = self

    output = '' 

    while node != None:

      output += str(node.val)

      output += " "

      node = node.next

    print(output)



  # Iterative Solution

  def reverseIteratively(self, head):

    # Implement this.



  # Recursive Solution      

  def reverseRecursively(self, head):

    # Implement this.



# Test Program

# Initialize the test list: 

testHead = ListNode(4)

node1 = ListNode(3)

testHead.next = node1

node2 = ListNode(2)

node1.next = node2

node3 = ListNode(1)

node2.next = node3

testTail = ListNode(0)

node3.next = testTail



print("Initial list: ")

testHead.printList()

# 4 3 2 1 0

testHead.reverseIteratively(testHead)

#testHead.reverseRecursively(testHead)

print("List after reversal: ")

testTail.printList()

# 0 1 2 3 4