【leetcode】(python)155. Min Stack最小栈

155. Min Stack Easy

Description

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.

题意

设计一个堆栈，实现其入栈、出栈、取顶及最小值操作。

解体思路

这题考察平时所用函数的具体实现，栈是特殊的列表，所以初始化时将其声明为 [] 类型，然后注意求最小值的过程，可以定义栈里每个元素为(x, cur_min)，表示当前元素和此栈的最小值。

code

class MinStack(object):

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

    def push(self, x):
        """
        :type x: int
        :rtype: None
        """
        pre_min = 2147483647 if len(self.stack) == 0 else self.stack[-1][1]
        cur_min = min(x, pre_min)
        self.stack.append((x, cur_min))
        

    def pop(self):
        """
        :rtype: None
        """
        self.stack.pop()
        

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

    def getMin(self):
        """
        :rtype: int
        """
        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()

又是过了半个月的刷题。。