大家好,我是毛毛。ヾ(´∀ ˋ)ノ
那就開始今天的解題吧~
Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.
Implement the MinStack
class:
MinStack()
initializes the stack object.void push(int val)
pushes the element val
onto the stack.void pop()
removes the element on the top of the stack.int top()
gets the top element of the stack.int getMin()
retrieves the minimum element in the stack.
You must implement a solution with O(1)
time complexity for each function.
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
Constraints:
-2^31 <= val <= 2^31 - 1
pop
, top
and getMin
operations will always be called on non-empty stacks.3 * 10^4
calls will be made to push
, pop
, top
, and getMin
.基本上就是要用陣列實作出stack的各種操作~
val
塞進stack中class MinStack:
def __init__(self):
self.stack = []
def push(self, val: int) -> None:
self.stack.append(val)
return None
def pop(self) -> None:
self.stack.pop(-1)
return None
def top(self) -> int:
return self.stack[-1]
def getMin(self) -> int:
return min(self.stack)
# Your MinStack object will be instantiated and called as such:
# obj = MinStack()
# obj.push(val)
# obj.pop()
# param_3 = obj.top()
# param_4 = obj.getMin()
今天就到這邊啦~
大家明天見