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.
My code:
class MinStack(object):
def __init__(self):
"""
initialize your data structure here.
"""
self.record=[]
def push(self, x):
"""
:type x: int
:rtype: void
"""
self.record.append(x)
def pop(self):
"""
:rtype: void
"""
self.record=self.record[:-1]
def top(self):
"""
:rtype: int
"""
return self.record[-1]
def getMin(self):
"""
:rtype: int
"""
return min(self.record)
# 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()Solution: https://leetcode.com/problems/min-stack/