Problem
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.
Solution
Use a second stack to keep track of the min value. Pay attention to the return value. Remember to return None when there is no result.
class MinStack:
def __init__(self):
"""
initialize your data structure here.
"""
self.stack = []
self.min = []
def push(self, x: int) -> None:
self.stack.append(x)
if not self.getMin() or self.getMin() >= x:
self.min.append(x)
def pop(self) -> None:
if self.stack:
x = self.stack.pop()
if x == self.min[-1]:
self.min.pop()
def top(self) -> int:
if self.stack:
return self.stack[-1]
else:
return None
def getMin(self) -> int:
if self.min:
return self.min[-1]
else:
return None