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 1:
Input
["MinStack","push","push","push","getMin","pop","top","getMin"]
[[],[-2],[0],[-3],[],[],[],[]]
Output
[null,null,null,null,-3,null,0,-2]
Explanation
MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin(); // return -3
minStack.pop();
minStack.top(); // return 0
minStack.getMin(); // return -2
Solution
我们在push的时候同时记录目前的最小值，如果要push的值比当前的最小值还小，那么把新的值存在最小值中，否则存上之前的最小值。我们用python的列表来模拟stack，stack中的元素是一个元祖，元祖的第一个元素为push的值，第二个为当前stack中的最小值。
class MinStack:
def __init__(self):
"""
initialize your data structure here.
"""
self.stack=[]
def push(self, x: int) -> None:
if len(self.stack)==0:
self.stack.append((x,x))
else:
self.stack.append((x,min(x,self.getMin())))
def pop(self) -> None:
self.stack.pop()
def top(self) -> int:
return self.stack[-1][0]
def getMin(self) -> int:
return self.stack[-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()
