Min Stack

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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.

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剑指offerP132的第21题，分析思路那里讲的很清楚，即利用额外的一个minStack来保存当前stack的最小值（在数据每次入栈的时候完成这一工作）。

代码如下：

class MinStack {    private Stack<Integer> stack = new Stack<Integer>();    //minStack用来记录当前时刻stack栈中的最小元素    private Stack<Integer> minStack = new Stack<Integer>();    public void push(int x) {        stack.push(x);        if(minStack.empty()){            minStack.push(x);        }else if(minStack.peek()<=x){            minStack.push(minStack.peek());        }else{            minStack.push(x);        }                    }    public void pop() {       stack.pop();       minStack.pop();    }    public int top() {        return stack.peek();    }    public int getMin() {        return minStack.peek();    }}

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