题解 | #数组遍历# 方法很多

import java.util.Scanner;
import java.util.Arrays;

public class Main {
    public static void main(String[] args) {
        int[] ary = new int[6];
        int max = 0;
        int min = 0;
        Scanner scanner = new Scanner(System.in);
     	for (int i = 0; i <ary.length ; i++) {
            ary[i]=scanner.nextInt();
            if( i == 0)
            {
                max = ary[i];
                min = ary[i];
            }
            else
            {
                if(max < ary[i])
                {
                    max = ary[i];
                }
                if(min > ary[i])
                {
                    min = ary[i];
                }

            }
        }

        //write your code here......
        max = ary[0];
        min = ary[0];
        for(int i=0;i<ary.length;i++)
        {
            max=Math.max(ary[i],max);
            min=Math.min(ary[i],min);
        }

        Arrays.sort(ary);//先排序->升序
        max = ary[ary.length-1];//最后一个最大
        min = ary[0];//第一个最小


        System.out.println(max+" "+min);
    }
}

遍历获得大小，直接比较；调用Math.max(a,b) Math.min(a,b)

import java.util.Arrays

Arrays.sort(); 排序也可以

方法太多了，都不熟练