L155_Min Stack.md

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.

####算法思路

在建立栈的时候，同时建立最小栈

最小栈的每一个输入的数值是当前栈的最小值

class MinStack(object):

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

    def push(self, x):
        """
        :type x: int
        :rtype: void
        """
        self.stack.append(x)
        if not self.min_stack:
            self.min_stack.append(x)
        else:
            self.min_stack.append(min(self.min_stack[-1],x))
        

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

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

    def getMin(self):
        """
        :rtype: int
        """
        return self.min_stack[-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()