Python 实现栈与队列

#基于Python2.7

#基于顺序表实现

#发现用Python写题时，没有像写C++时方便的STL可用，不过查阅资料之后发现用class实现也很简洁，不过效率应该不是很高

Python实现栈并使用：

class SStack():
    def __init__(self):
        self._elems=[]
    def is_empty(self):
        return self._elems==[]
    def top(self):
        #if self._elems==[]:
        #    raise StackUnderflow("in SStack.top()")
        return self._elems[-1]
    def push(self,e):
        self._elems.append(e)
    def pop(self):
        #if self._elems==[]:
        #    raise StackUnderflow("in SStack.pop()")
        return self._elems.pop()
S=SStack()
for i in range(3):
    S.push(i)
while not S.is_empty():
    temp=S.top()
    print temp
    S.pop()

Python实现队列并使用：

class SQueue():  
    def __init__(self,init_len=8):
        self._len=init_len
        self._elems=[0]*init_len
        self._head=0
        self._num=0
    def is_empty(self):
        return self._num==0
    def peek(self):
        #if self._num==0:
        #    raise QueueUnderflow
        return self._elems[self._head]
    def dequeue(self):
        #if self._num==0:
        #    raise QueueUnderflow
        e=self._elems[self._head]
        self._head=(self._head+1)%self._len
        self._num-=1
        return e
    def enqueue(self,e):
        if self._num==self._len:
            self.__extend()
        self._elems[(self._head+self._num)%self._len]=e
        self._num+=1
    def __extend(self):
        old._len=self._len
        self._len*=2
        new_elems=[0]*self._len
        for i in range(old_len):
            new_elems[i]=self._elems[(self._head+i)%old_len]
        self._elems,self._head=new_elems,0
Q=SQueue()
for i in range(5):
    Q.enqueue(i)
while not Q.is_empty():
    print Q.dequeue()
    Q.peek()