Leetcode: Min Stack

本文介绍了一个支持 push、pop、top 和获取最小元素的栈的设计,包括实现细节和代码示例。

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Question

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.


Solution

Analysis

refer to here.
Great Idea !

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



    def push(self, x):
        """
        :type x: int
        :rtype: nothing
        """
        self.stack.append(x)
        if self.minstack==[] or self.minstack[-1]>=x:
            self.minstack.append(x)

    def pop(self):
        """
        :rtype: nothing
        """
        if self.stack!=[]:
            elem = self.stack[-1]
            self.stack = self.stack[0:-1]
            if self.minstack!=[] and elem==self.minstack[-1]:
                self.minstack = self.minstack[0:-1]

    def top(self):
        """
        :rtype: int
        """
        if self.stack!=[]:
            return self.stack[-1]
        else:
            return None


    def getMin(self):
        """
        :rtype: int
        """
        if self.minstack!=[]:
            return self.minstack[-1]
        else:
            return None
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