Python程序：从链表中删除指定目标的最后一个出现位置

假设我们有一个单向链表和另一个值 called “target”，我们需要删除给定链表中 “target” 的最后一个出现位置。

因此，如果输入是 [5,4,2,6,5,2,3,2,4,5,4,7]，目标为5，那么输出将为 [5, 4, 2, 6, 5, 2, 3, 2, 4, 4, 7,]。

为了解决这个问题，我们将按照以下步骤进行：

head := node
k := null, prev := null
found := False
while node is not null, do
- if value of node is same as target, then
  - found := True
  - prev := k
- k := node
- node := next of node
if found is False, then
- return head
if prev is null, then
- return next of head
next of prev := next of the next of prev
return head

让我们看一下以下实现以更好地理解。

更多Python相关文章，请阅读：Python 教程

示例

class ListNode:
   def __init__(self, data, next = None):
      self.val = data
      self.next = next
def make_list(elements):
   head = ListNode(elements[0])
   for element in elements[1:]:
      ptr = head
      while ptr.next:
         ptr = ptr.next
   ptr.next = ListNode(element)
   return head
def print_list(head):
   ptr = head
   print('[', end = "")
   while ptr:
      print(ptr.val, end = ", ")
   ptr = ptr.next
   print(']')
class Solution:
   def solve(self, node, target):
      head = node
      k = None
      prev = None
      found = False
      while node:
         if node.val == target:
            found = True
         prev = k
         k = node
         node = node.next
         if found == False:
            return head
         if not prev:
            return head.next
            prev.next = prev.next.next
      return head
ob = Solution()
L = make_list([5,4,2,6,5,2,3,2,4,5,4,7])
target = 5
print_list(ob.solve(L, target))