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.
因为最小数可能被pop出,所以append需要额外存储次小数:
class MinStack(object):
def __init__(self):
"""
initialize your data structure here.
"""
self.stack=[]
def push(self, x):
"""
:type x: int
:rtype: void
"""
curMin=self.getMin()
if curMin==None or curMin>x:
curMin=x
self.stack.append((x, curMin))
def pop(self):
"""
:rtype: void
"""
self.stack.pop()
def top(self):
"""
:rtype: int
"""
return self.stack[-1][0]
def getMin(self):
"""
:rtype: int
"""
if not self.stack:
return None
return self.stack[-1][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()