用于查找链表长度的Python程序

编写一个函数来计算给定单链表中的节点数。

例如，对于链表 1->3->1->2->1，函数应该返回 5。

迭代解决方案：

1) Initialize count as 0 
2) Initialize a node pointer, current = head.
3) Do following while current is not NULL
     a) current = current -> next
     b) count++;
4) Return count 

以下是上述算法的迭代实现，用于查找给定单链表中的节点数。

Python

# A complete working Python program to 
# find the length of a Linked List
# iteratively
  
# Node class
class Node:
  
    # Function to initialize the node object
    def __init__(self, data):
  
        # Assign data
        self.data = data 
  
        # Initialize next as null
        self.next = None 
  
# Linked List class contains a Node object
class LinkedList:
  
    # Function to initialize head
    def __init__(self):
        self.head = None
  
    # This function is in LinkedList class. 
    # It inserts a new node at the beginning
    # of Linked List.
    def push(self, new_data):
  
        # 1 & 2: Allocate the Node &
        #        Put in the data
        new_node = Node(new_data)
  
        # 3. Make next of new Node as head
        new_node.next = self.head
  
        # 4. Move the head to point to the new Node
        self.head = new_node
  
    # This function counts number of nodes in 
    # Linked List iteratively, given 'node' 
    # as starting node.
    def getCount(self):
  
        # Initialise temp
        temp = self.head 
        count = 0 # Initialise count
  
        # Loop while end of linked list is 
        # not reached
        while (temp):
            count += 1
            temp = temp.next
        return count
  
# Code execution starts here
if __name__=='__main__':
    llist = LinkedList()
    llist.push(1)
    llist.push(3)
    llist.push(1)
    llist.push(2)
    llist.push(1)
    print ("Count of nodes is :",
            llist.getCount())

输出：

count of nodes is 5

递归解决方案：

int getCount(head)
1) If head is NULL, return 0.
2) Else return 1 + getCount(head->next)

以下是上述算法的递归实现，用于查找给定单链表中的节点数。

Python

# A complete working Python program to 
# find the length of a Linked List 
# recursively
  
# Node class
class Node:
    # Function to initialize the node object
    def __init__(self, data):
  
        # Assign data
        self.data = data  
        
        # Initialize next as null
        self.next = None  
  
# Linked List class contains a Node object
class LinkedList:
  
    # Function to initialize head
    def __init__(self):
        self.head = None
  
    # This function is in LinkedList class. 
    # It inserts a new node at the beginning 
    # of Linked List.
    def push(self, new_data):
  
        # 1 & 2: Allocate the Node &
        #        Put in the data
        new_node = Node(new_data)
  
        # 3. Make next of new Node as head
        new_node.next = self.head
  
        # 4. Move the head to point to the new Node
        self.head = new_node
  
    # This function counts number of nodes in 
    # Linked List recursively, given 'node' 
    # as starting node.
    def getCountRec(self, node):
  
        # Base case
        if (not node): 
            return 0
        else:
            return 1 + self.getCountRec(node.next)
  
    # A wrapper over getCountRec()
    def getCount(self):
       return self.getCountRec(self.head)
  
# Code execution starts here
if __name__=='__main__':
    llist = LinkedList()
    llist.push(1)
    llist.push(3)
    llist.push(1)
    llist.push(2)
    llist.push(1)
    print ("Count of nodes is :",
            llist.getCount())

输出：

Count of nodes is 5

有关详细信息，请参阅有关查找链表长度（迭代和递归）的完整文章！