Easy, Stack

Question:

设计一个支持push, pop, top, getMin的堆栈。时间复杂度为O(1)
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.

Solution

构造两个堆栈，一个储存实际数据，一个储存最小数。下面的解放更进一步，利用tuple将两个堆栈融合在一起。

class MinStack(object):

    def __init__(self):
        """
        initialize your data structure here.
        """
        self.stack = []
        
    def push(self, x):
        """
        :type x: int
        :rtype: void
        """
        curMin = self.getMin()
        if curMin == None or x < curMin:
            curMin = x
        self.stack.append((x,curMin))

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

        

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

        

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