[leetcode] 155. Min Stack @ python

原题

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.

解法

题意要求用固定时间获取最小值, 因此我们构建列表, 将元组(x, cur_min)放入列表, 每次push时将x与之前的最小值比较, 如果之前的最小值不存在或者x比之前的最小值更小, 那么x就是当前最小值.

Time: O(1)
Space: O(1)

代码

class MinStack(object):

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

    def push(self, x):
        """
        :type x: int
        :rtype: void
        """
        prev_min = self.getMin()
        if prev_min == None or x < prev_min:
            self.stack.append((x, x))
        else:
            self.stack.append((x, prev_min))
        

    def pop(self):
        """
        :rtype: void
        """
        self.stack.pop()

    def top(self):
        """
        :rtype: int
        """
        return self.stack[-1][0]

    def getMin(self):
        """
        :rtype: int
        """
        if len(self.stack) == 0:
            return None
        return self.stack[-1][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()