[155] Min Stack
https://leetcode.com/problems/min-stack/description/
- algorithms
- Easy (34.12%)
- Source Code: 155.min-stack.py
- Total Accepted: 286.9K
- Total Submissions: 787.3K
- Testcase Example: '["MinStack","push","push","push","getMin","pop","top","getMin"]\n[[],[-2],[0],[-3],[],[],[],[]]'
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
class MinStack(object):
def __init__(self):
"""
initialize your data structure here.
"""
self.stack = []
self.small_stack = []
def push(self, x):
"""
:type x: int
:rtype: None
"""
self.stack.append(x)
if not self.small_stack or self.small_stack[-1] >= x:
self.small_stack.append(x)
def pop(self):
"""
:rtype: None
"""
if self.small_stack[-1] == self.stack[-1]:
self.small_stack.pop(-1)
self.stack.pop(-1)
def top(self):
"""
:rtype: int
"""
return self.stack[-1]
def getMin(self):
"""
:rtype: int
"""
return self.small_stack[-1]
# Your MinStack object will be instantiated and called as such:
# ["MinStack","push","push","push","getMin","pop","getMin"]
# [[],[0],[1],[0],[],[],[]]
# obj = MinStack()
# obj.push(0)
# obj.push(1)
# obj.push(0)
# param_1 = obj.getMin()
# obj.pop()
# param_2 = obj.getMin()