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.
Python Solution:
class MinStack(object):
def __init__(self):
"""
initialize your data structure here.
"""
self.stack = []
self.min = 0
def push(self, x):
"""
:type x: int
:rtype: void
"""
if not self.stack:
self.stack.append(0)
self.min = x
else:
self.stack.append(x - self.min)
if x < self.min: self.min = x
def pop(self):
"""
:rtype: void
"""
if not self.stack: return
pp = self.stack.pop()
if pp < 0:
self.min -= pp
def top(self):
"""
:rtype: int
"""
if self.stack[-1] < 0:
return self.min
return self.min + self.stack[-1]
def getMin(self):
"""
:rtype: int
"""
return self.min
# 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()
Summary:
- beautiful.
- There is another solution, even more beautiful. Try to figure it out.
LeetCode: 155. Min Stack
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