Min Stack
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.
题目大意:
设计一个栈,支持在常数时间内push,pop,top,和取最小值。
push(x) -- 元素x压入栈
pop() -- 弹出栈顶元素
top() -- 获取栈顶元素
getMin() -- 获取栈中的最小值
解题思路:
“双栈法”,栈stack存储当前的所有元素,minStack存储栈中的最小元素。
在操作元素栈stack的同时,维护最小值栈minStack。
class MinStack(object):
def __init__(self):
"""
initialize your data structure here.
"""
self.node = []
self.minnode =[]
def push(self, x):
"""
:type x: int
:rtype: void
"""
self.node.append(x)
if self.minnode:
x = min(self.minnode[-1], x)
self.minnode.append(x)
def pop(self):
"""
:rtype: void
"""
self.minnode.pop()
self.node.pop()
def top(self):
"""
:rtype: int
"""
return self.node[-1]
def getMin(self):
"""
:rtype: int
"""
return self.minnode[-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()
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