正文:
编程语言:
python
题目:
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.
提示:
pop、top 和 getMin 操作总是在 非空栈 上调用。
来源:
来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/min-stack
编程代码:
(python)
class MinStack:
def __init__(self):
"""
initialize your data structure here.
"""
self.stack = []
self.MinStack = [inf]
def push(self, x: int) -> None:
self.stack.append(x)
self.MinStack.append(min(self.MinStack[-1],x))
def pop(self) -> None:
self.MinStack.pop()
self.stack.pop()
def top(self) -> int:
return self.stack[-1]
def getMin(self) -> int:
return self.MinStack[-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()