【leetcode】155. Min Stack

题目要求：

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.
• 即让你设计一个栈，支持push，pop，top和获取最小元素的操作
• 基本思想：借助一个辅助栈记为minStack，存储最小值。原来的栈记为stack。
• 当进行push操作时，如果stack为空，就把元素同时push进stack和minStack，如果stack不为空，将元素正常push进stack，然后获取minStack的栈顶元素，和即将入栈的元素进行比较，将较小者push进minStack
• 当进行pop操作时，两个栈都要pop
• 当进行top操作时，获取stack的栈顶元素即可
• 当要获取栈的最小值时，只要从minStack中获取栈顶元素即可

• class MinStack {    public Stack<Integer> stack  = new Stack<Integer>();    public Stack<Integer> minStack = new Stack<Integer>();    public void push(int x) {        if(stack.isEmpty())        {            stack.push(x);            minStack.push(x);        }else        {            stack.push(x);            int min = minStack.peek();            minStack.push(Math.min(x,min));        }    }    public void pop() {        stack.pop();        minStack.pop();    }    public int top() {        return stack.peek();    }    public int getMin() {        return minStack.peek();    }}