更快地找到数组中的最小值和最大值

Question

我想知道,编译器的哪种方式会更快.使用sort或循环方法显示最小值和最大值.对于编译器或程序员来说,哪种方法更快？

我将循环方法作为注释包含在内.

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

public class Day3 
{
    public static void main(String[] args) 
    {
        Scanner scan = new Scanner(System.in);
        System.out.println("Maximum and minimum value array");
        System.out.println("--------------------------------\n");

        //variable declaration
        int amount;
        int[] integerArray;

        //set amount
        System.out.print("Enter the amount of numbers to sort: ");
        amount = scan.nextInt();
        integerArray = new int[amount];

        //input loop
        System.out.print("Enter numbers: ");
        for(int i = 0; i < amount; i++)
        {
            integerArray[i] = scan.nextInt();
        }

        System.out.print("\nHere is your array: " + Arrays.toString(integerArray) + "\n");

        Arrays.sort(integerArray);
        System.out.println("Min value = " + integerArray[0]);
        System.out.println("Max value = " + integerArray[integerArray.length - 1]);
        System.out.println("Median value = " + integerArray[(integerArray.length -1) / 2]);

        /*
        int[] arr = new int[10];
        Scanner in = new Scanner(System.in);
        int i, min=0, max=0;
        for(i=0; i<=arr.length; i++)
        {
            System.out.print("Enter any number: ");
            arr[i] = in.nextInt();          
        }
        min = arr[0];
        for(i=0; i<=9; i++)
        {
            if(arr[i] > max)
            {
                max = arr[i];
            }
            if(arr[i] < min)
            {
                min = arr[i];
            }
        }
        System.out.println("Maximum is: " + max);
        System.out.println("Minimum is: " + min);
        */

    }
}

Answer 1

对于未排序的数据,循环遍历数组以查找最小值/最大值需要O(n)时间.对于排序它将是恒定时间(O(1))但是据我所知,这不是你的情况.

最快的排序算法在O(n*log(n))中工作,因此线性扫描(循环)是最快的选择.

此外,当编译器可以为您优化某些内容时,情况并非如此.