Leetcode +155 Min Stack

Leetcode +155 Min Stack"/>

Leetcode +155 Min Stack

Leetcode +155 Min Stack

题目描述

Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.

push(x) – Push element x onto stack.
pop() – Removes the element on top of the stack.
top() – Get the top element.
getMin() – Retrieve the minimum element in the stack.

Example:

MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin();   --> Returns -3.
minStack.pop();
minStack.top();      --> Returns 0.
minStack.getMin();   --> Returns -2.

来源：力扣（LeetCode）
链接：
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思路解析

class MinStack:def __init__(self):# 数据栈self.data = []# 辅助栈self.helper = []def push(self, x):self.data.append(x)if len(self.helper) == 0 or x <= self.helper[-1]:self.helper.append(x)else:self.helper.append(self.helper[-1])def pop(self):if self.data:self.helper.pop()return self.data.pop()def top(self):if self.data:return self.data[-1]def getMin(self):if self.helper:return self.helper[-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()