劍指offer刷題日記 棧 佇列 堆 雜湊表

棧05-兩個棧實現佇列（python）

#stack1入 stack2出

class

solution
:	def __init__
(self)
:		self.stack1=
self.stack2=
def push
(self,node)
:		self.stack1.
(node)
def pop
(self):if
len(self.stack1)
==0 or len
(self.stack2)==0
:return 
elif len
(self.stack2)==0
:while
len(self.stack1)
>0:
self.stack2.
(self.stack1.
pop(
))

#輔助棧minstack

class

solution
:	def __init__
(self)
:		self.stack=
self.minstack=
def push
(self,node)
:if not self.stack:
self.stack.
(node)
else
:if self.minstack[-1
]self.minstack.
(self.minstack[-1
])else
:				self.minstack.
(node)
def pop
(self)
:		self.stack.
pop(-1
)		self.minstack.
pop(-1
)	def top
(self)
:		self.stack.
pop(-1
)	def min
(self)
:		self.minstack.
pop(-1
)

#stack1入 stack2出

class

solution
:	def ispoporder
(self,pushv,popv)
:		stack=
for i in pushv:
stack.
(i)while stack and stack[-1
]==popv[0]
:				stack.
pop(
)				popv.
pop(0)
iflen
(stack)==0
:return true
else
:return false

#按空格拆分

class

solution
:	def reversesetence
(self,s)
:		stack=
[n for n in s.
split
(' ')]
stack.
reverse()
return''.
join
(stack)

class

solution
:	def maxwindows
(self,num,size)
:if not nums or size<=0:
return
queue=
res=
for i in
range
(len
(num)):
while queue and queue[0]
<=i-size:
queue.
pop(0)
while queue and nums[i]
>num[queue[-1
]]:				queue.
pop(-1
)			queue.
(i)if i>=size-1:
res.
(num[queue[0]
)return res

class

solution
:	def getleastnumbers_solution
(self,tinput,k)
:if not tinput or k>
len(tinput)
:return
tinput=self.
quicksort
(tinput)
return tinput[
:k]	def quicksort
(self,list)
:if not list:
return
pivot=list[0]
left=self.
quicksort
([x for x in list[1:
]if xright=self.
quicksort
([x for x in list[1:
]if x>=pivot)
return left+
[pivot]
+right

class

solution
:	def fistnotrepeating
(self,s)
:		dict=
for i in s:
dicts[i]
=dicts.
get(i,0)
+1for i in s:
if dicts[i]==1
:return i
return
-1

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