[155. Min Stack(https://leetcode.com/problems/min-stack/description/)
题目描述 设计一个支持push，pop，top等操作并且可以在O(1)时间内检索出最小元素的堆栈。 push(x)–将元素x插入栈中 pop()–移除栈顶元素 top()–得到栈顶元素 getMin()–得到栈中最小元素 例子 MinStack minStack = new MinStack(); minStack.push(-1); minStack.push(3); minStack.push(-4); minStack.getMin(); --> Returns -4. minStack.pop(); minStack.top(); --> Returns 3. minStack.getMin(); --> Returns -1. 思想 每次push进一个数时，辅助栈minStack存储截止当前位置的最小值。 class MinStack: def __init__(self): """ initialize your data structure here. """ self.value=[] self.stackmin=[] def push(self, x: int) -> None: self.value.append(x) if self.stackmin and self.stackmin[-1]<x: self.stackmin.append(self.stackmin[-1]) else: self.stackmin.append(x) def pop(self) -> None: self.value.pop() self.stackmin.pop() def top(self) -> int: return self.value[-1] def getMin(self) -> int: return self.stackmin[-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()