題目

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.

思路：

1. 這道題是design的題，以前只遇到過一次，design的題還是得多做
2. 里面有一個trick是怎么用常數時間去get（min）， python里面用tuple去存

Python

class MinStack(object):

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

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

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

    def top(self):
        """
        :rtype: int
        """
        if not self.s: return None
        return self.s[-1][0]

    def getMin(self):
        """
        :rtype: int
        """
        if not self.s: return None
        return self.s[-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()