设计一个支持 push ,pop ,top 操作,并能在常数时间内检索到最小元素的栈。
push(x) —— 将元素 x 推入栈中。
pop() —— 删除栈顶的元素。
top() —— 获取栈顶元素。
getMin() —— 检索栈中的最小元素。
示例:
输入:
["MinStack","push","push","push","getMin","pop","top","getMin"]
[[],[-2],[0],[-3],[],[],[],[]]
输出:
[null,null,null,null,-3,null,0,-2]
解释:
MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin(); --> 返回 -3.
minStack.pop();
minStack.top(); --> 返回 0.
minStack.getMin(); --> 返回 -2.
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/min-stack
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
import sys
class MinStack:
def __init__(self):
"""
initialize your data structure here.
"""
self.minList=[]
self.minHelp=[]
self.m=sys.maxsize
self.minHelp.append(self.m)
def push(self, x: int) -> None:
self.minList.append(x)
self.minHelp.append(min(self.minHelp[-1],x))
def pop(self) -> None:
self.minList.pop()
self.minHelp.pop()
def top(self) -> int:
return self.minList[-1]
def getMin(self) -> int:
return self.minHelp[-1]
class MinStack:
def __init__(self):
"""
initialize your data structure here.
"""
self.ms=[]
self.m=0
def push(self, x: int) -> None:
if self.ms:
self.m=min(self.ms[-1][1],x)
else:
self.m=x
self.ms.append([x,self.m])
def pop(self) -> None:
return self.ms.pop()
def top(self) -> int:
return self.ms[-1][0]
def min(self) -> int:
return self.ms[-1][1]