# coding=utf8'''題目：定義棧的數(shù)據(jù)結(jié)構(gòu)，請在該類型中實現(xiàn)一個能夠得到棧的最小元素的min函數(shù)。在該棧中，調(diào)用min、push及pop的時間復雜度都是O(1)。'''class Stack():  def __init__(self):    self.main_stack = []    # 輔助棧，每次次最小的元素壓入輔助棧    self.assist_stack = []    # 記錄棧中的最小元素    self._min = None  def min(self):    return self._min  def push(self, data):    self.main_stack.append(data)    if self._min is None:      self._min = data    else:      if data < self._min:        self._min = data    # 將最小的元素壓入輔助棧    self.assist_stack.append(self._min)  def pop(self):    if len(self.main_stack) == 0:      raise Exception('no data')    elif len(self.main_stack) == 1:      self.assist_stack.pop()      self._min = None      return self.main_stack.pop()    else:      self.assist_stack.pop()      self._min = self.assist_stack[-1]      return self.main_stack.pop()if __name__ == '__main__':  s = Stack()  s.push(3)  s.push(4)  s.push(2)  s.push(1)  print s.min()  s.pop()  s.pop()  print s.min()  s.pop()  print s.min()  s.pop()  print s.min()  s.pop()