原始題目

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.

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

題目分析

設計具有 push, pop, top 操作，並能在常數時間 O(n) 內查詢到最小元素的堆疊

解題過程

1. 建立一個 list min_stack紀錄堆疊最小值
2. 建立stack list 作為堆疊紀錄
3. push：把值放進堆疊後面，如果該值小於等於 min_stack 堆疊的最後一筆資料，則一併加入 min_stack 堆疊
4. pop：從 stack 移除最上面一筆值，如果該值等於 min_stack 最後一筆，一併從堆疊移除
5. top：取得堆疊最上面一個元素stack[-1]
6. getMin：取得最小值，也就是 min_stack 的最後一個元素min_stack[-1]

class MinStack:
    def __init__(self):
        self.stack = []
        self.min_stack = []

    def push(self, val: int) -> None:
        self.stack.append(val)
        if not self.min_stack or val <= self.min_stack[-1]: 
            self.min_stack.append(val)

    def pop(self) -> None:
        if self.stack.pop() == self.min_stack[-1]:
            self.min_stack.pop()
    
    def top(self) -> int:
        return self.stack[-1]
    
    def getMin(self) -> int:
        return self.min_stack[-1]

# 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()

結果