题目
Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.
- push(x) – Push element x onto stack.
- pop() – Removes the element on top of the stack.
- top() – Get the top element.
- getMin() – Retrieve the minimum element in the stack.
解题思路
额外维护一个min_stack，使其是非严格递降的，即push时如果是小于等于栈顶元素则加入，pop时，如果发现pop的元素和min_stack的栈顶元素相同，则同时pop两个栈
代码
class MinStack:
def __init__(self):
"""
initialize your data structure here.
"""
self.stack = []
self.min_stack = [float('inf')]
def push(self, x: int) -> None:
self.stack.append(x)
if x <= self.min_stack[-1]:
self.min_stack.append(x)
def pop(self) -> None:
if self.stack[-1] == self.min_stack[-1]:
self.min_stack.pop()
self.stack.pop()
def top(self) -> int:
return self.stack[-1]
def getMin(self) -> int:
return self.min_stack[-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.getMin()
