leetcode155. Min Stack

原题链接https://leetcode.com/problems/min-stack/

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 1:

Input
["MinStack","push","push","push","getMin","pop","top","getMin"]
[[],[-2],[0],[-3],[],[],[],[]]

Output
[null,null,null,null,-3,null,0,-2]

Explanation
MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin(); // return -3
minStack.pop();
minStack.top(); // return 0
minStack.getMin(); // return -2
创建一个最小值栈专门储存最小值，随着栈的更新更新最小值栈。

class MinStack:

    def __init__(self):
        """
        initialize your data structure here.
        """
        self.stack = []
        self.minstack = []

    def push(self, x: int) -> None:
        self.stack.append(x)
        if self.minstack:
            if self.minstack[-1] >= x:
                self.minstack.append(x)
        else:
            self.minstack.append(x)

    def pop(self) -> None:
        if self.stack:
            a = self.stack.pop()
            
            if a == self.minstack[-1]:
                self.minstack.pop()

    def top(self) -> int:
        if self.stack:
            return self.stack[-1]
        return []

    def getMin(self) -> int:
        if self.minstack:
            return self.minstack[-1]
        return []