18 刪除鍊錶的節點

class

node
(object):
''' 節點類
'''def__init__
(self, data)
: self.num = data
self.
next
=none
class
deletenode
(object):
''' 實現刪除指定節點功能
'''defdelete_node
(self, head, delnode)
:if head ==
none
or delnode ==
none
:return
if head == delnode and delnode.
next
isnone
:# 頭節點
head.num =
none
if delnode.
next
==none
:# 尾節點
while head:
if head.
next
== delnode:
head.
next
=none
head =head.
next
else
:# 中間節點
delnode.num = delnode.
next
.num
delnode.
next
= delnode.
next
.next
# 刪除重複節點
defdeleteduplication
(self, head)
:if head is
none
:return
while head and head.
next
isnot
none
:if head.num == head.
next
.num:
head.
next
= head.
next
.next
head = head.
next
class
printnode()
:'''
輸出指定節點為起始節點的鍊錶
'''defprint_node
(self, node)
: res_list =
while node:
str(node.num)
) node = node.
next
print
('->'
.join(res_list)
)if __name__ ==
'__main__'
: node1 = node(90)
# node2 = node(90)
# node3 = node(77)
# node4 = node(77)
# node5 = node(23)
# node1.next = node2
# node1.next.next = node3
# node1.next.next.next = node4
# node1.next.next.next.next = node5
# 原鍊錶節點
print
('init single linknode is:'
) printnode = printnode(
) printnode.print_node(node1)
# 刪除鍊錶中某個節點
delete = deletenode(
)# delete.delete_node(node1, node3)
# print('after delete node,the single linknode is:')
# printnode.print_node(node1)
# 刪除重複節點
delete.deleteduplication(node1)
print
('after delete duplicationnode,the single linknode is:'
) printnode.print_node(node1)

18 刪除鍊錶節點

給定單向鍊錶的頭指標和乙個要刪除的節點的值，定義乙個函式刪除該節點。返回刪除後的鍊錶的頭節點。注意 此題對比原題有改動 示例 1 輸入 head 4,5,1,9 val 5 輸出 4,1,9 解釋 給定你鍊錶中值為 5 的第二個節點，那麼在呼叫了你的函式之後，該鍊錶應變為 4 1 9.示例 2 輸入...

18 刪除鍊錶的節點

在乙個排序的鍊錶中，存在重複的結點，請刪除該鍊錶中重複的結點，重複的結點不保留，返回煉表頭指標。例如，鍊錶1 2 3 3 4 4 5 處理後為 1 2 5 1.首先新增乙個頭節點，以方便碰到第乙個，第二個節點就相同的情況 2.設定 pre last 指標，pre指標指向當前確定不重複的那個節點，而l...

JZ 18 刪除鍊錶的節點（鍊錶）

設當前節點指標 cur curcu r 和前驅節點指標 pre prepr e，遍歷鍊錶找到目標節點後執行 pre nex t cu r.ne xtpre.next cur.next pre.ne xt c ur.n ext 即可刪除 cur curcu r 節點 演算法流程 definition ...