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: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。
一解
使用两个列表:
- 第一个列表:存储元素,模拟出入栈
- 第二个列表:存储当前栈的最小元素
class MinStack:
def __init__(self):
"""
initialize your data structure here.
"""
self.min_stack = []
self.stack = []
def push(self, val: int) -> None:
self.stack.append(val)
if not self.min_stack or val <= self.min_stack[-1]:
self.min_stack.append(val)
def pop(self) -> None:
val = self.stack.pop()
if self.min_stack and val == self.min_stack[-1]:
self.min_stack.pop()
def top(self) -> int:
if not self.stack:
return None
return self.stack[-1]
def getMin(self) -> int:
if not self.min_stack:
return None
return self.min_stack[-1]
# Your MinStack object will be instantiated and called as such:
# obj = MinStack()
# obj.push(val)
# obj.pop()
# param_3 = obj.top()
# param_4 = obj.getMin()
- 时间复杂度:O(1)
- 空间复杂度:O(1)
二解
思路与一解相同,向 min ,可减少判断:
class MinStack:
def __init__(self):
"""
initialize your data structure here.
"""
self.min_stack = [math.inf]
self.stack = []
def push(self, val: int) -> None:
self.stack.append(val)
self.min_stack.append(min(val, self.min_stack[-1]))
def pop(self) -> None:
self.stack.pop()
self.min_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(val)
# obj.pop()
# param_3 = obj.top()
# param_4 = obj.getMin()
- 时间复杂度:O(1)
- 空间复杂度:O(1)