前言

大家好，我是毛毛。ヾ(´∀ ˋ)ﾉ
那就開始今天的解題吧~

Question

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
• Methods pop, top and getMin operations will always be called on non-empty stacks.
• At most 3 * 10^4 calls will be made to push, pop, top, and getMin.

Think

基本上就是要用陣列實作出stack的各種操作~

• push
  • 把val塞進stack中
• pop
  • 把stack最後一個放進去的值刪掉
• getMin
  • 找出stack中最小值
• top
  • 回傳目前最後一個放進去的值

Code

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()

Submission

今天就到這邊啦~
大家明天見