JonnyF--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.

解题思路：

     这道题也很简单，只是单纯的实现一个栈的结构。只要理解了栈的结构和功能就很好做了。

class MinStack:
    def __init__(self):
        self.items = []
        self.minitems = []

    # @param x, an integer
    # @return an integer
    def push(self, x):
        self.items.append(x)
        if not self.minitems:
            self.minitems.append(x)
        elif x <= min(self.minitems):
            self.minitems.append(x)
        return x

    # @return nothing
    def pop(self):
        if not self.items:
            return
        stop = self.items[len(self.items) - 1]
        mstop = self.minitems[len(self.minitems) - 1]
        if stop == mstop:
            self.minitems.pop()
        self.items.pop()

    # @return an integer
    def top(self):
        if not self.items:
            return 0
        return self.items[len(self.items) - 1]

    # @return an integer
    def getMin(self):
        if not self.minitems:
            return 0
        return self.minitems[len(self.minitems) - 1]