155. 最小栈

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:

  1. 第一个列表:存储元素,模拟出入栈
  2. 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)
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