155. Min Stack-最小栈

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 1:
Input
["MinStack","push","push","push","getMin","pop","top","getMin"]
[[],[-2],[0],[-3],[],[],[],[]]

Output
[null,null,null,null,-3,null,0,-2]

Explanation
MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin(); // return -3
minStack.pop();
minStack.top(); // return 0
minStack.getMin(); // return -2

Constraints:

• Methods pop, top and getMin operations will always be called on non-empty stacks.

Solution in python:

class MinStack:

    def __init__(self):
        """
        initialize your data structure here.
        """
        self.stack = []
        self.min_element = float('-inf')

    def push(self, x: int) -> None:
        if not self.stack or x < self.min_element:
            self.min_element = x
        self.stack.append(x)

    def pop(self) -> None:
        if len(self.stack) == 1:
            self.min_element = float('-inf')
        elif self.stack[-1] == self.min_element:
            self.min_element = min(self.stack[:-1])
        self.stack.pop()

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

    def getMin(self) -> int:
        return self.min_element

# 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()