包含min函数的栈
定义栈的数据结构，请在该类型中实现一个能够得到栈的最小元素的 min 函数在该栈中，调用 min、push 及 pop 的时间复杂度都是 O(1)。
示例:
MinStack minStack = new MinStack(); minStack.push(-2); minStack.push(0); minStack.push(-3); minStack.min(); --> 返回 -3. minStack.pop(); minStack.top(); --> 返回 0. minStack.min(); --> 返回 -2.
提示：
- 各函数的调用总次数不超过 20000 次
示例代码：
class MinStack(object):
def __init__(self):
"""
initialize your data structure here.
"""
self.A, self.B = [], []
def push(self, x):
"""
:type x: int
:rtype: None
"""
self.A.append(x)
if not self.B or self.B[-1] >= x:
self.B.append(x)
def pop(self):
"""
:rtype: None
"""
tmp = self.A.pop()
if tmp == self.B[-1]:
self.B.pop()
def top(self):
"""
:rtype: int
"""
return self.A[-1]
def min(self):
"""
:rtype: int
"""
return self.B[-1]
# Your MinStack object will be instantiated and called as such:
# obj = MinStack()
# obj.push(x)
# obj.pop()
# param_3 = obj.top()
# param_4 = obj.min()
示例代码2：
class MinStack(object):
def __init__(self):
"""
initialize your data structure here.
"""
self.stack = []
def push(self, x):
"""
:type x: int
:rtype: None
"""
if not self.stack:
self.stack.append((x, x))
else:
min = self.min()
if x < min:
self.stack.append((x, x))
else:
self.stack.append((x, min))
def pop(self):
"""
:rtype: None
"""
self.stack.pop()
def top(self):
"""
:rtype: int
"""
return self.stack[-1][0]
def min(self):
"""
:rtype: int
"""
return self.stack[-1][1]
# Your MinStack object will be instantiated and called as such:
# obj = MinStack()
# obj.push(x)
# obj.pop()
# param_3 = obj.top()
# param_4 = obj.min()
