Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.
Implement the MinStack
class:
MinStack()
initializes the stack object.void push(int val)
pushes the elementval
onto the stack.void pop()
removes the element on the top of the stack.int top()
gets the top element of the stack.int getMin()
retrieves 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:
-231 <= val <= 231 - 1
- Methods
pop
,top
andgetMin
operations will always be called on non-empty stacks. - At most
3 * 104
calls will be made topush
,pop
,top
, andgetMin
.
class ListNode:
def __init__(self, x, min = None):
self.val = x
self.next = None
self.min = min
class MinStack:
def __init__(self):
self.head = None
self.min = None
def push(self, value: int) -> None:
if not self.head:
self.head = ListNode(value, value)
self.min = value
else:
min = value if value < self.min else self.min
self.min = min
node = ListNode(value, min)
node.next = self.head
#self.head.next = node
self.head = node
def pop(self) -> None:
if self.head is not None:
self.head = self.head.next
if self.head is not None:
self.min = self.head.min
else:
self.min = None
def top(self) -> int:
if self.head is not None:
return self.head.val
def getMin(self) -> int:
return self.min
# Your MinStack object will be instantiated and called as such:
# obj = MinStack()
# obj.push(val)
# obj.pop()
# param_3 = obj.top()
# param_4 = obj.getMin()