1.10 Given a particular node deletes the node pointer claim

The node to delete only the single case for a pointer to a linked list of nodes

Subject description:

Suppose a given list 1-> 2-> 3-> 4-> 5-> 6-> 7, pointer to the first element 5, 5 requires the node to delete, delete the list becomes 1-> 2- > 3-> 4-> 6-> 7

Analysis of ideas:

1. If this node is the last node of the list, you can not delete this node.

2. If the last node is not the node linked list, by copying the data to its successor node in the current node, then delete the successor node of methods.

   Code:

   # --coding:utf-8--
   """
   @Author : 图南
   @Software: PyCharm
   @Time : 2019/9/7 19:46
   """
   class Node:
   def init(self, data=None, next=None):
   self.data = data
   self.next = next

   def printLink(head):
   if head is None or head.next is None:
   return
   cur = head.next
   while cur != None:
   print(cur.data, end=" ")
   cur = cur.next
   print()

   def conLink(nums, n):
   nums = list(map(int, nums.split(' ')))
   n = int(n)
   if len(nums) == 0 or n == 0:
   return
   p = None
   head = Node()
   cur = head
   for i in range(1, len(nums)+1):
   node = Node(nums[i-1])
   cur.next = node
   cur = node
   if i == n:
   p = cur
   return head, p

   def deleteP(p):
   if p.next is None:
   return False
   p.data = p.next.data
   p.next = p.next.next
   return True

   IF name == ' main ':
   the nums = INPUT ( 'list:')
   n-INPUT = ( 'nodes:')
   head, P = conLink (the nums, n-)
   Print ( 'before deleting:')
   PrintLink (head)
   F deleteP = (the p-)
   IF f:
   Print ( 'after deleting:')
   PrintLink (head)
   the else:
   Print ( 'can not be deleted!')

   operation result: