155. 最小栈

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155. 最小栈

官方链接

https://leetcode-cn.com/problems/min-stack/

设计一个支持 push，pop，top 操作，并能在常数时间内检索到最小元素的栈。

push(x) -- 将元素 x 推入栈中。 pop() -- 删除栈顶的元素。 top() -- 获取栈顶元素。 getMin() -- 检索栈中的最小元素。

示例:

MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin();   --> 返回 -3.
minStack.pop();
minStack.top();      --> 返回 0.
minStack.getMin();   --> 返回 -2.

解法一

数据栈 + 辅助栈

class MinStack {

    private Stack<Integer> dataStack;
    private Stack<Integer> minStack;

    /** initialize your data structure here. */
    public MinStack() {
        dataStack = new Stack<>();
        minStack = new Stack<>();
    }

    public void push(int x) {
        dataStack.push(x);
        if(minStack.isEmpty() || x <= minStack.peek()) minStack.push(x);
    }

    public void pop() {
        int x = dataStack.pop();
        if(x == minStack.peek()) {
            minStack.pop();
        }
    }

    // public void pop() {
    //     if(stack.peek().equals(minStack.peek())) minStack.pop();
    //     stack.pop();
    // }

    public int top() {
        return dataStack.peek();
    }

    public int getMin() {
        return minStack.peek();
    }
}

/**
 * Your MinStack object will be instantiated and called as such:
 * MinStack obj = new MinStack();
 * obj.push(x);
 * obj.pop();
 * int param_3 = obj.top();
 * int param_4 = obj.getMin();
 */

解法二

class MinStack:

    def __init__(self):
        """
        initialize your data structure here.
        """
        self.stack = []


    def push(self, x: int) -> None:
        self.stack.append(x)

    def pop(self) -> None:
        return self.stack.pop()

    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(x)
# obj.pop()
# param_3 = obj.top()
# param_4 = obj.getMin()