栈（2）

原题：

/** * Created by gouthamvidyapradhan on 08/03/2017. * Design a stack that supports push, pop, top, and retrieving the minimum element in constant time. * <p> * 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. * Example: * MinStack minStack = new MinStack(); * minStack.push(-2); * minStack.push(0); * minStack.push(-3); * minStack.getMin();   --> Returns -3. * minStack.pop(); * minStack.top();      --> Returns 0. * minStack.getMin();   --> Returns -2. */

答案：

public class MinStack {    class Node {        int value, min;        Node(int value, int min) {            this.value = value;            this.min = min;        }    }    private Stack<Node> stack = new Stack<>();    /**     * Main method     *     * @param args     * @throws Exception     */    public static void main(String[] args) throws Exception {        MinStack minStack = new MinStack();        minStack.push(-2);        minStack.push(0);        minStack.push(-3);        System.out.println(minStack.getMin());        minStack.pop();        System.out.println(minStack.top());        System.out.println(minStack.getMin());    }    public MinStack() {    }    public void push(int x) {        Node node;        if (!stack.isEmpty()) {            Node top = stack.peek();            node = new Node(x, Math.min(top.min, x));        } else {            node = new Node(x, x);        }        stack.push(node);    }    public void pop() {        stack.pop();    }    public int top() {        return stack.peek().value;    }    public int getMin() {        return stack.peek().min;    }}