LeetCode Min Stack

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Problem

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.

即自定义一个栈，除了支持栈的基本操作外，支持一个额外的getMin返回当前栈内元素最小值的操作。

Python 实现


# 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.

# author li.hzh

class MinStack:

    __data = []
    __min_list = []

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

    def push(self, x):
        """
        :type x: int
        :rtype: void
        """
        if not self.__data or x < self.__min_list[-1]:
            self.__min_list.append(x)
        else:
            self.__min_list.append(self.__min_list[-1])
        self.__data.append(x)

    def pop(self):
        """
        :rtype: void
        """
        self.__data.pop()
        self.__min_list.pop()

    def top(self):
        """
        :rtype: int
        """
        return self.__data[-1]

    def getMin(self):
        """
        :rtype: int
        """
        return self.__min_list[-1]


# Your MinStack object will be instantiated and called as such:
obj = MinStack()
obj.push(-1)
param_3 = obj.top()
param_4 = obj.getMin()
print(param_3)
print(param_4)

分析

单纯功能很简单，但是其实显然这题隐含了一个效率要求。如果，每次getMin都实时计算最小值的话，解法就超时了。因此，采用空间换时间的策略，用一个额外的数组保存每次操作后，栈内的最小值。该数组的最后一个元素，即为当前的最小值。

Thanks a lot.