剑指offer_21

题意：实现带有min函数的栈
思路：用两个栈，一个num栈，一个min栈，当压入一个数字时候，若是比min栈中最小值大，那么min栈依然压入min栈栈顶的数字，否则，压入当前待压入的值。
代码：

package MianShiTi_21;

import java.util.Stack;
public class MianShiTi_21 {
    static Stack<Integer> numStack = new Stack<>();
    static Stack<Integer> minStack = new Stack<>();
    public static void push(int node) {
        if(minStack.empty()){
            minStack.push(node);
            numStack.push(node);
        }else{
            int tempNode = minStack.peek();
            if(node >= tempNode){
                minStack.push(tempNode);
                numStack.push(node);
            }else{
                minStack.push(node);
                numStack.push(node);
            }
        }
    }

    public static void pop() {
        if(!numStack.isEmpty() && !minStack.isEmpty()){
            numStack.pop();
            minStack.pop();
        }
    }

    public static int min() {
        if(minStack.isEmpty()){
            return 0;
        }
        return minStack.peek();
    }

    public static void main(String[] args) {
        push(3);
        System.out.println(min()+"a");
        push(4);
        System.out.println(min()+"b");
        push(5);
        System.out.println(min()+"c");
        push(1);
        System.out.println(min()+"d");
        push(2);
        System.out.println(min()+"e");
        pop();
        System.out.println(min()+"f");
    }
}