文件: root - text - article - 2019 - 12 - min-stack.txt
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标签: 每日算法题, 算法, 数据结构, 面试题, Daily Interview Problem, Data Structures and Algorithms, Computer Programming, Python, | 英文 | 主页 | 类别: 计算机科学 | 1229 次阅读, 17801 次搜索 | 137 个单词
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Hi, here's your problem today. This problem was recently asked by Uber:
Design a simple 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.
Note: be sure that pop() and top() handle being called on an empty stack.
标签: 每日算法题, 算法, 数据结构, 面试题, Daily Interview Problem, Data Structures and Algorithms, Computer Programming, Python, | 英文 | 主页 | 类别: 计算机科学 | 1229 次阅读, 17801 次搜索 | 137 个单词 Design a simple 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.
Note: be sure that pop() and top() handle being called on an empty stack.
class minStack(object):
def __init__(self):
# Fill this in.
def push(self, x):
# Fill this in.
def pop(self):
# Fill this in.
def top(self):
# Fill this in.
def getMin(self):
# Fill this in.
x = minStack()
x.push(-2)
x.push(0)
x.push(-3)
print(x.getMin())
# -3
x.pop()
print(x.top())
# 0
print(x.getMin())
# -2
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