Min Stack

题目地址：https://leetcode.com/problems/min-stack/

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.

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.

堆栈几个操作本身的实现问题不是很大，用LinkedList提供的方法基本就能搞定了，但是这里多了一个getMin方法，如果是每次调用getMin的时候都要从栈顶扫描到栈底去找最小值，那么这样是十分耗时的，那么我们就应该想办法在元素入栈的时候就想办法保存下来。

我们可以再设这样一个栈，这个栈根据上一个栈的push操作，如果出现更小的数字，则minStack执行push，如果上一个栈执行pop且pop出来的值等于minStack栈顶的值，那么minStack也执行pop操作，这样的话getMin的具体实现就是minStack.getFirst()啦，时间复杂度为O(1)：

public class MinStack {    /** initialize your data structure here. */    LinkedList<Integer> stack = new LinkedList<>();    LinkedList<Integer> minStack = new LinkedList<>();    public MinStack() {    }    public void push(int x) {        stack.push(x);        if (minStack.size() == 0)            minStack.push(x);        else {            if (minStack.getFirst() >= x)                minStack.push(x);        }    }    public void pop() {        int x = stack.pop();        if (x == minStack.getFirst()){            minStack.pop();        }    }    public int top() {        return stack.getFirst();    }    public int getMin() {        return minStack.getFirst();    }    public static void main(String[] args) {        MinStack stack = new MinStack();        stack.push(1);        stack.push(2);        stack.push(-3);        System.out.println(stack.getMin());        stack.pop();        System.out.println(stack.top());        stack.pop();        System.out.println(stack.top());    }}/** * 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(); */