题目
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.
思路：
- 这道题是design的题，以前只遇到过一次，design的题还是得多做
- 里面有一个trick是怎么用常数时间去get（min）， python里面用tuple去存
Python
class MinStack(object):
def __init__(self):
"""
initialize your data structure here.
"""
self.s = []
def push(self, x):
"""
:type x: int
:rtype: void
"""
curmin = self.getMin()
if curmin == None or x < curmin:
curmin = x
self.s.append((x,curmin))
def pop(self):
"""
:rtype: void
"""
self.s.pop()
def top(self):
"""
:rtype: int
"""
if not self.s: return None
return self.s[-1][0]
def getMin(self):
"""
:rtype: int
"""
if not self.s: return None
return self.s[-1][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()
