155. Min Stack
Read the full problem statement on LeetCode.
Difficulty: medium Acceptance: 56% Topics: Stack, Design
View full problem on LeetCode Reading material
Reference solution (spoiler · python)
# Time: O(n)
# Space: O(1)
class MinStack(object):
def __init__(self):
self.min = None
self.stack = []
# @param x, an integer
# @return an integer
def push(self, x):
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
# @return nothing
def pop(self):
x = self.stack.pop()
if x < 0:
self.min = self.min - x
# @return an integer
def top(self):
x = self.stack[-1]
if x > 0:
return x + self.min
else:
return self.min
# @return an integer
def getMin(self):
return self.min
# Time: O(n)
# Space: O(n)
class MinStack2(object):
def __init__(self):
self.stack, self.minStack = [], []
# @param x, an integer
# @return an integer
def push(self, x):
self.stack.append(x)
if len(self.minStack):
if x < self.minStack[-1][0]:
self.minStack.append([x, 1])
elif x == self.minStack[-1][0]:
self.minStack[-1][1] += 1
else:
self.minStack.append([x, 1])
# @return nothing
def pop(self):
x = self.stack.pop()
if x == self.minStack[-1][0]:
self.minStack[-1][1] -= 1
if self.minStack[-1][1] == 0:
self.minStack.pop()
# @return an integer
def top(self):
return self.stack[-1]
# @return an integer
def getMin(self):
return self.minStack[-1][0]
# time: O(1)
# space: O(n)
class MinStack3(object):
def __init__(self):
self.stack = []
def push(self, x):
if self.stack:
current_min = min(x, self.stack[-1][0])
self.stack.append((current_min, x))
else:
self.stack.append((x, x))
def pop(self):
return self.stack.pop()[1]
def top(self):
return self.stack[-1][1]
def getMin(self):
return self.stack[-1][0]
Solution from kamyu104/LeetCode-Solutions · MIT
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