155. 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:

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.

来自 <https://leetcode.com/problems/min-stack/description/>

思路1：使用python自带的列表实现，然后min就用官方的min函数

 1 class MinStack(object):
 2 
 3     def __init__(self):
 4         """
 5         initialize your data structure here.
 6         """
 7         self.l = []
 8 
 9     def push(self, x):
10         """
11         :type x: int
12         :rtype: void
13         """
14         self.l.append(x)
15 
16     def pop(self):
17         """
18         :rtype: void
19         """
20         del(self.l[-1])
21 
22     def top(self):
23         """
24         :rtype: int
25         """
26         return self.l[-1]
27 
28     def getMin(self):
29         """
30         :rtype: int
31         """
32         return min(self.l)

思路2：发现思路1速度不理想，并且排行榜上两个块比较集中，查看后发现更好的那一种优化了min实现，在添加元素时就做了判断。

class MinStack(object):

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

    def push(self, x):
        """
        :type x: int
        :rtype: void
        """
        self.l.append(x)
        if not len(self.min_l) or x <= self.min_l[-1]:
            self.min_l.append(x)

    def pop(self):
        """
        :rtype: void
        """
        if self.l[-1] == self.min_l[-1]:
            del (self.min_l[-1])
        del (self.l[-1])

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

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