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()