如何合併兩個有序鍊錶

【albb面試題】

題目：

把兩個有序的鍊錶合併成乙個有序的鍊錶，例如 1 -> 5 -> 6 和 2 -> 3 -> 7 -> 8，合併之後變成了 1 -> 2 -> 3 -> 5 -> 6 -> 7 -> 8。

# 合併兩個有序鍊錶

class
node
:def
__init__
(self, data)
:        self.data = data
self.
next
=none
class
link
:def
__init__
(self, head=
none
, tail=
none):
self.head = head
self.tail = tail
def(self, x)
:        node = node(x)
if self.head is
none
:            self.head = node
else
:            self.tail.
next
= node
self.tail = node
link_a = link(
)link_b = link(
)for i in
range(1
,8):
if i &1:
continue
print
("合併前："
)print
("link_a: "
, end=
" ")
p = link_a.head
while p:
print
(p.data, end=
"\t"
)    p = p.
next
print()
p = link_b.head
print
("link_b: "
, end=
" ")
while p:
print
(p.data, end=
"\t"
)    p = p.
next
defmerge
(link1, link2)
:# 合併函式
p1 = link1.head
p2 = link2.head
ifnot p1 or
not p2:
# 為空
return link1 if p1 else link2
while p1 and p2:
# 不為空
minnode = p1 if p1.data <= p2.data else p2
if p1 is link1.head and p2 is link2.head:
head = minnode
tail = minnode
else
:            tail.
next
= minnode
tail = tail.
next
if p1.data < p2.data:
p1 = p1.
next
else
:            p2 = p2.
next
if p1:
tail.
next
= p1
else
:        tail.
next
= p2
newlink = link(
)# 生成新的鍊錶
newlink.head = head
newlink.tail = tail
return newlink
a = merge(link_a, link_b)
p = a.head
print()
print
("合併後： "
, end="")
while p:
print
(p.data, end=
"\t"
)    p = p.
next

如何合併兩個有序鍊錶

已知兩個鍊錶head1和head2各自有序 例如公升序排列 請把他們合併成乙個鍊錶，要求合併後的鍊錶依然有序。class lnode def init self self.data none self.next none def constructlist start 方法功能 構造鍊錶 param...

合併兩個有序鍊錶

鍊錶的題目總是讓我很惆悵。動輒就會runtime error。比如這題，額外用了乙個節點的空間來儲存頭節點。我很不情願多用這個空間，不過貌似不行。貌似不行，實際可行，見附錄。把頭節點提出迴圈 實現類 class solution else if l1 null p next l1 if l2 nul...

合併兩個有序鍊錶

三個指標乙個儲存la鍊錶 乙個儲存lb鍊錶，乙個指向新的鍊錶。鍊錶的插入，兩個指標，乙個是head，乙個指向head後面的鏈，新插入的元素位於head後面。執行該 自己外加上class類。static class node public static void main string args st...