LeetCode:Min Stack

问题描述：

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.

思路：

刚开始想到的方法是遍历整个栈，将最小值返回，最后想到以前的一道题目和这到题目类似，果断采取了两个栈，一个放最小值。有一个地方是需要特别注意的：

if(stack.peek().equals(minstack.peek()))这里用到了一个函数equals，目的是比较二者大小。

JAVA代码：

class MinStack {
    Stack<Integer> stack = new Stack<>();
    Stack<Integer> minstack = new Stack<>();
    public void push(int x) {
        if(minstack.isEmpty() || x <= minstack.peek())
            minstack.push(x);
        stack.push(x);
    }

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

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

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