Easy, Stack
Question:
设计一个支持push, pop, top, getMin的堆栈。时间复杂度为O(1)
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.
Solution
构造两个堆栈，一个储存实际数据，一个储存最小数。下面的解放更进一步，利用tuple将两个堆栈融合在一起。
class MinStack(object):
def __init__(self):
"""
initialize your data structure here.
"""
self.stack = []
def push(self, x):
"""
:type x: int
:rtype: void
"""
curMin = self.getMin()
if curMin == None or x < curMin:
curMin = x
self.stack.append((x,curMin))
def pop(self):
"""
:rtype: void
"""
self.stack.pop()
def top(self):
"""
:rtype: int
"""
if self.stack:
return self.stack[-1][0]
def getMin(self):
"""
:rtype: int
"""
if self.stack:
return self.stack[-1][1]
