Leetcode 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.

 思路：用两个stack。一个用于维护最小值。例如push [10, 5, 6, 3, 1, 9, 6, 4]。minStack = [10, 5, 3, 1]。

如果stack[-1]不等于minStack[-1]，我们只要pop stack就可以，不用管minStack。

class MinStack(object):

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

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

    def pop(self):
        """
        :rtype: void
        """
        if self.stack[-1] == self.minStack[-1]:
            self.minStack.pop()
        return self.stack.pop()


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