從尾到頭列印鍊錶
鍊錶中倒數第k個結點
class solution:
def findkthtotail(self, head, k):
# write code here
if not head:
return none
pfast, pslow = head, head
for i in range(k):
if pfast == none:
return none
pfast = pfast.next
while pfast:
pfast = pfast.next
pslow = pslow.next
return pslow
反轉鍊錶
class solution:
# 返回listnode
def reverselist(self, phead):
# write code here
if not phead:
return none
pre = none
while phead:
next_node = phead.next # 提前儲存當前節點的下一節點
phead.next = pre # 翻轉，當前節點的下一節點是其上一節點
pre = phead # 上一節點向後移動
phead = next_node # 當前節點向後移動
return pre
合併兩個排序的鍊錶
class solution:
# 返回合併後列表
def merge(self, phead1, phead2):
# write code here
if not phead1:
return phead2
if not phead2:
return phead1
mergehead = none
if phead1.val <= phead2.val:
mergehead = phead1
mergehead.next = self.merge(phead1.next, phead2)
elif phead1.val > phead2.val:
mergehead = phead2
mergehead.next = self.merge(phead1, phead2.next)
return mergehead
複雜鍊錶的複製
class solution:
# 返回 randomlistnode
def clone(self, phead):
# write code here
if phead is none:
return none
# copy node set next
cur_node = phead
while cur_node:
next_node = cur_node.next
copy_node = randomlistnode(cur_node.label)
cur_node.next = copy_node
copy_node.next = next_node
cur_node = next_node
# set random
cur_node = phead
while cur_node:
next_node = cur_node.next
if cur_node.random is none:
next_node.random = none
else:
next_node.random = cur_node.random.next
cur_node = next_node.next
# split
new_head = phead.next
cur_node = phead
while cur_node:
next_node = cur_node.next
cur_node.next = next_node.next
if next_node.next is none:
next_node.next = none
else:
next_node.next = next_node.next.next
cur_node = cur_node.next
return new_head
兩個鍊錶的第乙個公共結點
鍊錶中環的入口結點
刪除鍊錶中重複的結點
class solution:
def deleteduplication(self, phead):
# write code here
if not phead:
return none
prenode = listnode(none)
prenode.next = phead
last = prenode
cur = phead
while cur and cur.next:
if cur.val == cur.next.val:
val = cur.val
while cur and cur.val == val:
cur = cur.next
last.next = cur
else:
last = cur
cur = cur.next
return prenode.next
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刷題筆記 鍊錶相關
輸入乙個鍊錶，反轉鍊錶後，輸出新鍊錶的表頭。分析 初始時，將 pren 指向 nullptr，curn 指向鍊錶首部，提前記錄好鍊錶的下乙個節點。為了反轉鍊錶，要將 curn 的 next 指標指向 pren。class solution return curn 輸入乙個鍊錶，輸出該鍊錶中倒數第k個...