题目描述

定义栈的数据结构，请在该类型中实现一个能够得到栈中所含最小元素的min函数（时间复杂度应为O（1））。

注意：保证测试中不会当栈为空的时候，对栈调用pop()或者min()或者top()方法。

解题思路一

public class Solution{
    private Stack<Integer> stack = new Stack<Integer>();
    private Stack<Integer> helpStack = new Stack<Integer>();//辅助栈
    private Integer minValue = Integer.MAX_VALUE;//记录当前栈最小值
    public void push(int node) {
        if(stack.empty()){
            minValue = node;
        }else{
            if(minValue > node)
                minValue = node;
        }
        stack.push(node);
    }

    public void pop() { //弹出，空间效率O（N）
        Integer value = stack.pop();
        if(value == minValue){//如果弹出最小值，借助辅助栈重新找到最小值
            minValue = Integer.MAX_VALUE;
            while (!stack.empty()){
                value = stack.pop();
                if(value < minValue)
                    minValue = value;
                helpStack.push(value);
            }
            //倒插回去
            while (!helpStack.empty()){
                stack.push(helpStack.pop());
            }
        }
    }

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

    public int min() {
        return minValue;
    }
}

解题思路二

public class Solution{
    private Stack<Integer> stack = new Stack<Integer>();
    private Stack<Integer> littleStack = new Stack<Integer>();  //存放最小值的栈
    private Integer minValue = Integer.MAX_VALUE;
    public void push(int node) {
        if(littleStack.empty()){
            littleStack.push(node);
        }else{
            if(node <= littleStack.peek())
                littleStack.push(node);
        }
        stack.push(node);
    }

    public void pop() {
        if(stack.empty())
            return ;//栈空无法弹栈
        int popValue = stack.pop();
        if(popValue == littleStack.peek())
            littleStack.pop();  //同时弹出两栈的最小值
    }

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

    public int min() {
        return littleStack.peek();
    }
}